Calculus

Step 1: Implicit Differentiation.
Differentiate the given equation implicitly with respect to : Using the chain rule, we get:
Step 2: Find the Slope of the Normal.
The slope of the tangent is , so the slope of the normal is the negative reciprocal of the tangent slope:
Step 3: Equation of the Normal.
Using the slope of the normal and the point on the curve, the equation of the normal can be written in point-slope form: Using the angle between the normal and the x-axis, we arrive at the desired equation:
Step 4: Conclusion.
Thus, the equation of the normal is:

Step 1: Let the expression be .
Using the chain rule, we differentiate with respect to .

Step 2: Differentiating .
Using the quotient rule:
Step 3: Simplifying the expression.
Substitute this into the derivative: Now, simplify the denominator. We know that: Thus, the derivative is:
Final Answer:

Step 1: Differentiate the equation implicitly.
The given equation is: Differentiate both sides with respect to : Simplifying: Solving for :
Step 2: Find the slope of the tangent at the point .
Substitute and : So, the slope of the tangent at the point is .
Step 3: Find the slope of the normal.
The slope of the normal is the negative reciprocal of the slope of the tangent:
Step 4: Use the point-slope form to find the equation of the normal.
Using the point-slope form , where and the point , we get: Simplifying: Thus, the equation of the normal at the point is:

Using the identity : Separate the terms: The first term evaluates to: The second term evaluates to 0 because is an odd function. Thus:

The function can be written as: The left-hand derivative at : The right-hand derivative at : Since , is not differentiable at .

Let . Taking the natural logarithm on both sides: Differentiating both sides with respect to : Multiply by to get :

Step 1: Understanding the Concept:
This is an optimization problem that requires the use of differential calculus. We need to express the volume of the box as a function of the side length 'x' of the cut-out squares, and then find the value of 'x' that maximizes this volume.
Step 2: Key Formula or Approach:
1. Determine the dimensions (length, width, height) of the resulting open box in terms of x.
2. Write the volume function .
3. Find the derivative and set it to zero to find critical points.
4. Use the second derivative test, , to confirm that the critical point corresponds to a maximum.
Step 3: Detailed Explanation:
The original sheet has dimensions L = 8 m and W = 3 m.
When squares of side x are cut from each corner, the sheet is folded up. The dimensions of the resulting box will be:
- Height: - Length: - Width: The volume of the box is . For the dimensions to be positive, we must have , , and . The valid domain for x is .
Now, we find the first derivative to find critical points: Set : Divide by 4: Factor the quadratic equation: The possible values for x are or .
From our domain , the only valid solution is .
To confirm this is a maximum, we use the second derivative test: Since , the volume is maximum at .
Calculate the maximum volume: Step 4: Final Answer:
The maximum volume of the box is cubic meters.

Step 1: Understanding the Concept:
This integral is in the form . This form often simplifies to a standard integral type , which has a direct solution. The strategy is to simplify the trigonometric function and see if it can be expressed as a sum of a function and its derivative.
Step 2: Key Formula or Approach:
The key integration formula is:
We will also use the trigonometric double-angle identities:
1.
2.
Step 3: Detailed Explanation or Calculation:
Let the integral be . First, we simplify the trigonometric part of the integrand:
Factor out 2 from the numerator and cancel with the denominator:
Now, split the fraction into two parts:
Simplify each part:
Now, substitute this simplified expression back into the integral:
Let's check if this fits the form .
If we let , then its derivative is .
So, the integrand is indeed in the form , where .
Using the formula:
Step 4: Final Answer:
The result of the integration is .

Step 1: Understanding the Concept:
A function is considered increasing on an interval if its first derivative is positive on that interval. So, to show that is increasing for , we need to find its derivative, , and show that for all .
Step 2: Key Formula or Approach:
The condition for an increasing function is:
The power rule for differentiation states that .
Step 3: Detailed Explanation or Calculation:
First, we find the function given:
Next, we differentiate with respect to to find :
Now, we need to check the condition for the function to be increasing, which is .
We are given the condition that .
If , then multiplying by a positive constant (14) will not change the inequality sign.
So, for all .
Step 4: Final Answer:
Since the first derivative is positive for all , the function is an increasing function when .

Step 1: Understanding the Concept:
This problem involves second-order differentiation. The goal is to find the first and second derivatives of the given function and then show that they satisfy the given differential equation.
Step 2: Key Formula or Approach:
We will use the chain rule for the first derivative and the product rule for the second derivative. 1. Find .
2. Rearrange the equation to avoid fractions, which simplifies the next differentiation.
3. Differentiate again with respect to using the product rule.
4. Rearrange the resulting equation to match the required form.
Step 3: Detailed Explanation:
We are given the function: Differentiating with respect to using the chain rule: To avoid using the quotient rule for the next step, multiply both sides by : Now, differentiate both sides again with respect to . We use the product rule on the left-hand side (LHS).
To eliminate the fraction on the right-hand side, multiply the entire equation by : This is the required expression. Hence, it is shown. Step 4: Final Answer:
By differentiating the function twice and performing algebraic manipulations, we have shown that .

Step 1: Understanding the Concept:
This is a classic problem in definite integrals. It requires the use of the properties of definite integrals, particularly the property . This allows us to create a second related integral which, when combined with the first, leads to a simpler expression that can be evaluated.
Step 2: Key Formula or Approach:
1. Let the given integral be . So, .
2. Apply the property with .
3. This will create a second equation for in terms of .
4. Add the two equations for . This will combine the logarithms into .
5. Use the identity to simplify the logarithm.
6. Split the resulting integral and solve for .
Step 3: Detailed Explanation:
Let .
Using the property : Since , we have: Now, add equations (1) and (2): Using the logarithm property : To use the double angle formula, multiply and divide by 2 inside the logarithm: Using the property : Let's evaluate the second integral first: Now consider the first integral, . Let , so or . The limits of integration change from and . Using the property if . Here , . , so the property applies. Substituting everything back into the equation for : Solving for I: Step 4: Final Answer:
We have proven that .

Step 1: Understanding the Concept:
This problem requires evaluating a definite integral. The integrand involves a square root of a trigonometric function, which can be simplified using a half-angle or double-angle identity.
Step 2: Key Formula or Approach:
We will use the trigonometric identity for the cosine of a double angle: By letting , this identity becomes: Step 3: Detailed Explanation:
Let the integral be I. Using the identity from Step 2, we can simplify the expression under the square root: The presence of the absolute value means we must consider the sign of over the interval . For , we have , where .
For , we have , where .
Therefore, we must split the integral at : Now, we evaluate each integral: For the first part: For the second part: Adding the results from both parts: Step 4: Final Answer:
We have shown that the value of the definite integral is 1.

Step 1: Understanding the Concept:
This is an integral of a square root of a quadratic expression. The standard technique is to complete the square for the quadratic expression inside the root to transform it into the form , , or . This allows the use of a standard integration formula.
Step 2: Key Formula or Approach:
1. Complete the square for the quadratic .
2. Rewrite the integral in the standard form .
3. Apply the standard integration formula: Step 3: Detailed Explanation:
First, complete the square for the expression inside the square root: So, the integral becomes: This is now in the form , where:

Now, apply the standard formula: Substitute back and : Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
This is an integral of a rational function. Since the degree of the numerator is less than the degree of the denominator, we can solve it using the method of partial fraction decomposition. This involves factoring the denominator and expressing the rational function as a sum of simpler fractions.
Step 2: Key Formula or Approach:
1. Factor the denominator polynomial.
2. Set up the partial fraction expansion based on the factors.
3. Solve for the unknown coefficients.
4. Integrate the resulting simpler fractions.
Step 3: Detailed Explanation or Calculation:
1. Factor the Denominator:
Let the denominator be . We can factor it by grouping:
The denominator has a repeated linear factor and a distinct linear factor .
2. Partial Fraction Decomposition:
We can write the integrand as:
To find the coefficients A, B, and C, multiply both sides by the denominator:
Now, substitute strategic values for x:
- For : .
- For : .
- For : .
Substitute the values of B and C:
3. Integrate:
Now, we can rewrite the integral as:
Step 4: Final Answer:
The solution is:
This can also be written using logarithm properties as:

Step 1: Understanding the Concept:
This problem involves evaluating a definite integral. The key to solving it is to simplify the integrand using a trigonometric identity before performing the integration. We must also be careful with the square root, as .
Step 2: Key Formula or Approach:
We will use the half-angle trigonometric identity: After substitution, the integral will involve , which needs to be handled by splitting the integral over intervals where is positive and negative.
Step 3: Detailed Explanation:
Let the integral be I. Using the identity , we get: Now, we need to consider the sign of in the interval .
For , , so .
For , , so .
We split the integral at : Now, we evaluate each part: Adding the results of the two parts: Step 4: Final Answer:
We have shown that the value of the integral is 2. Hence proved.

Step 1: Understanding the Concept:
The "slope" of a vector function at a particular point refers to the tangent vector at that point. The tangent vector is found by taking the derivative of the vector function with respect to its parameter (in this case, ).
Step 2: Key Formula or Approach:
The tangent vector (or slope) is given by the derivative .
We need to:
1. Differentiate the vector function with respect to .
2. Evaluate the resulting derivative vector at .

Step 3: Detailed Explanation:
The given vector function is: We differentiate each component of the vector function with respect to : The derivatives of the constant components are zero: This is the general expression for the tangent vector. Now, we evaluate it at : Step 4: Final Answer:
The slope of the vector function at is . Therefore, the correct option is (i).

Step 1: Understanding the Concept:
A function is continuous at a point if three conditions are met: is defined, exists, and . For the given piecewise function, we need to check continuity for and for .
Step 2: Key Formula or Approach:
For the case , we will use the Squeeze Theorem (or Sandwich Theorem) to evaluate the limit. The theorem states that if for all x in an open interval containing c (except possibly at c itself), and if , then .
Step 3: Detailed Explanation:
Case 1: For
The function is defined as . The functions , , and are all continuous on their respective domains. Since , the composition is continuous, and the product is also continuous. So, is continuous for all .
Case 2: For
We need to check if . From the function definition, we know that .
Now, let's evaluate the limit: We know that the sine function is bounded between -1 and 1, regardless of its input. Multiply the entire inequality by . Since , the direction of the inequalities does not change. Now, we take the limit of the bounding functions as : Since both the lower and upper bounds approach 0, by the Squeeze Theorem, the function in the middle must also approach 0. We have found that and we know . Since , the function is continuous at .
Step 4: Final Answer:
Since the function is continuous for all and also continuous at , we can conclude that the function is continuous for all real numbers. Hence proved.

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The graphical method involves plotting the constraints to identify the feasible region, finding the coordinates of the corner points of this region, and then evaluating the objective function Z at these corner points. The minimum (or maximum) value of Z will occur at one of these points.
Step 2: Detailed Explanation:
1. Graph the Constraints:
We treat the inequalities as equations to draw the boundary lines.
Line 1: . It passes through (10, 0) and (0, 5).
Line 2: . It passes through (5, 0) and (0, 15).
Line 3: (the y-axis).
Line 4: (the x-axis).
The inequalities and restrict the feasible region to the first quadrant. For the other two inequalities, shading towards the origin (0,0) satisfies them.
2. Identify the Feasible Region and Corner Points:
The feasible region is the polygon formed by the intersection of these half-planes. The corner points (vertices) are:
O (0, 0): The origin.
A (5, 0): The x-intercept of the line .
C (0, 5): The y-intercept of the line .
B: The intersection of lines and .
To find point B, we solve the system of equations: From (ii), . Substitute this into (i): Substitute back into : So, point B is (4, 3).
3. Evaluate Z at each Corner Point:
The objective function is Z = 3x + 2y. At O(0, 0): At A(5, 0): At B(4, 3): At C(0, 5): Step 3: Final Answer:
Comparing the values of Z, the minimum value is 0, which occurs at the corner point O(0, 0).

Step 1: Understanding the Concept:
The problem describes the motion of a car with its distance from the start point given as a function of time. The car stopping at point Q means its velocity becomes zero at that point in time. We need to use calculus to find the velocity from the distance function.
Step 2: Key Formula or Approach:
1. Express the distance as a polynomial of .
2. Velocity is the first derivative of distance with respect to time: .
3. Set the velocity to find the time when the car stops.
4. Substitute this value of back into the distance equation to find the total distance covered.
Step 3: Detailed Explanation:
The distance function is given by: To find the velocity, we differentiate with respect to : The car stops at point Q, so its velocity is zero. We set : This gives two possible times: and .
corresponds to the starting point P. Therefore, the time required for the car to reach point Q is seconds.
Now, we find the distance between P and Q by substituting into the distance equation: The distance between P and Q is metres.
Step 4: Final Answer:
The time required to reach point Q is 4 seconds, and the distance between P and Q is metres.

Step 1: Understanding the Concept:
The function is of the form , where both the base and the exponent are functions of x. This type of function is best differentiated using logarithmic differentiation.
Step 2: Key Formula or Approach:
The process of logarithmic differentiation involves: 1. Let . 2. Take the natural logarithm (ln) of both sides. 3. Use the logarithm property to simplify. 4. Differentiate both sides implicitly with respect to x, using the product rule on the right side. 5. Solve for .
Step 3: Detailed Explanation:
Let the given function be: Taking the natural logarithm of both sides: Using the power rule for logarithms: Now, differentiate both sides with respect to x. For the left side, we use the chain rule. For the right side, we use the product rule. To find , multiply both sides by y: Finally, substitute back the original expression for y: Step 4: Final Answer:
The derivative of is .

Step 1: Understanding the Concept:
This is an optimization problem where we need to find the semi-vertical angle of a cone that maximizes its volume, given that its slant height is constant. We will use calculus, specifically the first and second derivative tests, to find the maximum.
Step 2: Key Formula or Approach:
Let be the given slant height, be the height, be the radius, and be the semi-vertical angle of the cone.
The volume of the cone is given by the formula:
The relationships between , , , and are:
We need to express the volume as a function of a single variable, , and then find the value of for which is maximum.
Step 3: Detailed Explanation:
First, substitute the expressions for and in terms of and into the volume formula:
Since is a constant, to maximize , we need to maximize the function .
Now, we differentiate with respect to :
Using the product rule , let and .
So,
For maximum or minimum volume, we set :
Since and for a cone, , . Thus, we must have:
Dividing both sides by (which is non-zero for ):
This gives the critical point .
To confirm this is a maximum, we use the second derivative test. It's often easier to check the sign of the first derivative around the critical point. Or, we can analyze the derivative of . Let's find the second derivative of V.
At , we have . Substituting this into the bracket:
Since , , so .
Therefore, at , which confirms that the volume is maximum at this angle.
Step 4: Final Answer:
The volume of the cone is maximized when the semi-vertical angle satisfies .
Hence, the semi-vertical angle for the cone of maximum volume is .

Step 1: Understanding the Concept:
The area of a region bounded by a curve can be found using a definite integral. The ellipse given by is symmetric with respect to both the x-axis and the y-axis. We can calculate the area of the region in the first quadrant and then multiply it by 4 to get the total area of the ellipse.
Step 2: Key Formula or Approach:
The area of a region bounded by a curve , the x-axis, and the lines and is given by:
First, we need to express in terms of from the equation of the ellipse.
From , we get:
For the first quadrant, is positive, so we take . The limits of integration for in the first quadrant are from 0 to .
Step 3: Detailed Explanation:
The total area of the ellipse is 4 times the area of the part in the first quadrant.
To evaluate this integral, we use the trigonometric substitution .
Then .
We also need to change the limits of integration:
When , .
When , .
Substituting these into the integral:
Using the identity :
Now, use the identity :
Step 4: Final Answer:
The area of the region bounded by the ellipse is .

Step 1: Understanding the Concept:
This is a related rates problem from the application of derivatives. We are given the rate of change of the volume of a cone and a relationship between its height and radius. We need to find the rate of change of its height at a specific instant.
Step 2: Key Formula or Approach:
1. Volume of a cone: .
2. Identify the given quantities: and .
3. Express the volume solely in terms of the variable whose rate we want to find (in this case, ).
4. Differentiate the volume equation with respect to time .
5. Substitute the known values to solve for the unknown rate .
Step 3: Detailed Explanation:
We are given:
- Rate of change of volume, .
- Relationship between height and radius : . This implies .
- We need to find when cm.
The formula for the volume of a cone is .
To find , we should express as a function of only. Substitute into the volume formula:
Now, differentiate both sides with respect to time : Using the chain rule: Now, substitute the given values: and . Solve for : Step 4: Final Answer:
The height of the sand cone is increasing at a rate of cm/second.

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The goal is to find the maximum value of a linear objective function subject to a set of linear inequalities (constraints). The graphical method involves plotting the constraints to find a feasible region and then testing the corner points of this region in the objective function.
Step 2: Key Formula or Approach:
1. Convert the inequalities into equations to plot the boundary lines.
2. Identify the feasible region that satisfies all constraints.
3. Determine the coordinates of the corner points (vertices) of the feasible region.
4. Evaluate the objective function at each corner point.
5. The maximum value of will be the largest value among those calculated.
Step 3: Detailed Explanation:
The constraints are:




Let's plot the lines:
- (passes through (1,0) and (0,1))
- (passes through the origin (0,0) at a 45-degree angle)
- (the y-axis)
- (the x-axis)
The feasible region is the area bounded by these lines that satisfies all the inequalities. It is a triangle with vertices at the intersection points of these lines.
Let's find the corner points:
- Point O: Intersection of and . The coordinates are .
- Point A: Intersection of and .
Substituting into the first equation: . So, .
The coordinates are .
- Point B: Intersection of and .
Substituting : .
The coordinates are .
Now, evaluate the objective function at these corner points:
- At O(0, 0):
- At A(0.5, 0.5):
- At B(1, 0):
Step 4: Final Answer:
Comparing the values of Z, the maximum value is 1.5, which occurs at the point .
Maximum .

Step 1: Understanding the Concept:
This question asks for the evaluation of a standard indefinite integral. The integrand is of a specific form whose result is a well-known formula in calculus.
Step 2: Key Formula or Approach:
The integral is a standard form:
where is the constant of integration.
This formula can be derived using trigonometric substitution by setting .
Step 3: Detailed Explanation:
This is a direct application of a standard integration formula.
Given the integral: By comparing this with the standard formula , where , we can directly write the result.
Step 4: Final Answer:
The evaluation of the integral is .

Step 1: Understanding the Concept:
A function is said to be continuous at a point if three conditions are met:
1. is defined.
2. The limit of the function as approaches , , exists.
3. The limit equals the function's value: .
Step 2: Detailed Explanation:
The given function is . We need to test its continuity at .
Condition 1: Check if is defined.
Substitute into the function:
Since , we get:
The function is defined at .
Condition 2: Check if the limit exists.
We need to evaluate .
We can apply the limit to each term separately:
The limit exists.
Condition 3: Compare the limit and function value.
From our calculations, we have:
Since , all three conditions for continuity are met.
Step 3: Final Answer:
The function is continuous at .

Step 1: Understanding the Concept:
This problem involves integrating a trigonometric function. The strategy is to simplify the integrand and then apply standard integration formulas.
Step 2: Key Formula or Approach:
The key is to use the distributive property to expand the integrand and then use the following standard integrals:
1.
2.
Step 3: Detailed Explanation:
First, expand the expression inside the integral:
Now, we can split the integral into two parts using the sum rule for integration:
Apply the standard integration formulas to each part:
For the first part: .
For the second part: .
Combining the results and adding the constant of integration, :
Step 4: Final Answer:
The evaluation of the integral is .

Step 1: Understanding the Concept:
The function is of the form , where both the base and the exponent are functions of .
This type of function is best differentiated using logarithmic differentiation.
Step 2: Key Formula or Approach:
The process of logarithmic differentiation involves these steps:
1. Let .
2. Take the natural logarithm ( ) of both sides: .
3. Differentiate both sides with respect to using the product rule and chain rule.
4. Solve for and substitute the original expression for .
The product rule states that .
Step 3: Detailed Explanation:
Let the given function be .
Taking the natural logarithm on both sides, we get:
Using the logarithm property , we have:
Now, we differentiate both sides with respect to . The left side requires the chain rule, and the right side requires the product rule.
To find , we multiply both sides by :
Finally, substitute back into the equation.
Step 4: Final Answer:
The derivative of with respect to is .

Step 1: Understanding the Concept:
The problem asks for the total area enclosed by an ellipse given in its standard form. This can be found directly using the standard formula for the area of an ellipse.
Step 2: Key Formula or Approach:
The standard equation of an ellipse centered at the origin is .
The area (A) enclosed by such an ellipse is given by the formula: where and are the lengths of the semi-major and semi-minor axes, respectively.
Step 3: Detailed Explanation:
The given equation of the ellipse is: By comparing this with the standard form , we can identify the values of and .
Now, we apply the formula for the area of the ellipse: Substituting the values of and : Step 4: Final Answer:
The area of the region bounded by the ellipse is square units.

Step 1: Understanding the Concept:
This is an optimization problem using calculus (applications of derivatives). We are given a cone with a fixed total surface area and we need to find the semi-vertical angle that maximizes its volume. The key is to express the volume as a function of a single variable (either radius, height, or the semi-vertical angle) and then use differentiation to find the maximum.
Step 2: Key Formula or Approach:
1. Let be the radius, the height, the slant height, and the semi-vertical angle of the cone.
2. Write the formulas for total surface area and volume .
3. Use the geometric relations: , , .
4. Since S is constant, express one variable (e.g., ) in terms of another (e.g., ) from the surface area formula.
5. Substitute this into the volume formula to get V as a function of a single variable. To make differentiation easier, it's better to work with .
6. Differentiate with respect to the chosen variable, set the derivative to zero to find the condition for maximum volume.
7. Relate this condition back to the semi-vertical angle .
Step 3: Detailed Explanation:
Let S be the constant total surface area. .
From this, we can express the slant height in terms of : The volume of the cone is . We also know .
To simplify the differentiation, we will maximize instead of V. Substitute the expression for : Using the difference of squares : Let . Then .
To find the maximum, we differentiate Z with respect to and set it to zero: Setting : Since , we can divide by : This is the condition for maximum volume. Now we relate this back to the semi-vertical angle .
Substitute back into this condition: The semi-vertical angle is defined by .
Substituting : (One can verify this is a maximum by checking the second derivative, which will be negative).
Step 4: Final Answer:
The semi-vertical angle of the cone with a given surface area and maximum volume is .

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The goal is to find the minimum value of a linear objective function Z, subject to a set of linear inequalities called constraints. The graphical method involves plotting these constraints to find the feasible region and then evaluating the objective function at the corner points (vertices) of this region. The optimal solution (minimum or maximum) will occur at one of these vertices.
Step 2: Key Formula or Approach:
1. Convert the inequalities into equations to plot the lines.
2. Graph the lines on a 2D plane.
3. Identify the feasible region, which is the area that satisfies all the given constraints simultaneously.
4. Determine the coordinates of the corner points of the feasible region.
5. Evaluate the objective function Z = 200x + 500y at each corner point.
6. The smallest value of Z will be the minimum value.
Step 3: Detailed Explanation:
The constraints are: 1. Graph the lines:
Line 1: . It passes through (10, 0) and (0, 5). The region is the area on and above this line.
Line 2: . It passes through (8, 0) and (0, 6). The region is the area on and below this line.
The constraints restrict the feasible region to the first quadrant.
2. Find the feasible region and corner points:
The feasible region is the area in the first quadrant that is above the line and below the line . The vertices (corner points) of this region are:
Point A: Intersection of and . . So, A = (0, 6).
Point B: Intersection of and . . So, B = (0, 5).
Point C: Intersection of and .
From , we have . Substitute this into the second equation: Substitute back into : So, C = (4, 3).
3. Evaluate Z at corner points:
The objective function is Z = 200x + 500y.
At A(0, 6): .
At B(0, 5): .
At C(4, 3): .
4. Find the minimum value:
Comparing the values of Z, the minimum value is 2300.
Step 4: Final Answer:
The minimum value of Z is 2300, which occurs at the point (x, y) = (4, 3).

Step 1: Understanding the Concept:
To determine the intervals where a function is increasing or decreasing, we use the first derivative test. A differentiable function is: - Increasing on an interval if its first derivative, , is positive ( ) for all x in that interval. - Decreasing on an interval if its first derivative, , is negative ( ) for all x in that interval.
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, .
2. Determine the sign of in the interval .
3. Determine the sign of in the interval .
4. Conclude whether the function is increasing or decreasing based on the sign of the derivative in each interval.
Step 3: Detailed Explanation:
The given function is . The domain of this function requires , which is true for .
1. Find the first derivative, :
Using the chain rule, . 2. Analyze the sign of in :
The interval corresponds to the first quadrant.
In the first quadrant, both and are positive.
Therefore, is positive for all .
Since , the function is increasing in the interval .
3. Analyze the sign of in :
The interval corresponds to the second quadrant.
In the second quadrant, is positive, but is negative.
Therefore, is negative for all .
Since , the function is decreasing in the interval .
Step 4: Final Answer:
We have shown that in and in . Therefore, the function is increasing in the first interval and decreasing in the second.

Step 1: Understanding the Concept:
This problem involves an infinite power tower. The key to solving such problems is to recognize the self-similar nature of the expression. The expression in the exponent is the same as the original expression for y. This allows us to write a simpler, implicit equation for y, which can then be differentiated using logarithmic differentiation.
Step 2: Key Formula or Approach:
1. Rewrite the infinite tower in a finite form: .
2. Take the natural logarithm of both sides to bring the exponent down: .
3. Differentiate both sides of the equation implicitly with respect to .
4. Solve the resulting equation for .
5. Rearrange the expression to match the required form.
Step 3: Detailed Explanation:
Given the function: .
Due to the infinite nature of the tower, we can write: Take the natural logarithm on both sides: Using the logarithm property , we get: Now, differentiate both sides with respect to , using the product rule on the right side: Now, we need to solve for . Group the terms containing on one side: Factor out : Isolate : Finally, multiply both sides by to get the desired form: Step 4: Final Answer:
The relation has been proven.

Step 1: Understanding the Concept:
The problem asks for the derivative of an inverse trigonometric function. The expression inside the function is a hint to use a trigonometric substitution to simplify the function before differentiating. The expression is identical in form to the triple angle formula for tangent.
Step 2: Key Formula or Approach:
1. Recognize the trigonometric identity: .
2. Use the substitution , which implies .
3. Use the given domain of to find the range of .
4. Simplify the expression for using the identity and the property (within the principal value range).
5. Differentiate the simplified expression for with respect to .
Step 3: Detailed Explanation or Calculation:
Let . Then .
The given domain for is .
Substituting , we get .
This implies .
Now, substitute into the expression for : Using the triple angle identity, this simplifies to: To simplify this further, we must check if lies within the principal value range of , which is .
From our range for , , we can find the range for : Since is within the principal value range, we can write: Now, substitute back : Finally, differentiate with respect to : Step 4: Final Answer:
The value of is .

Step 1: Understanding the Concept:
This is a standard indefinite integral of a rational function. The standard method to solve it from first principles is using partial fraction decomposition. Alternatively, one can state the well-known formula directly. We will show the derivation using partial fractions.
Step 2: Key Formula or Approach:
1. Factor the denominator: .
2. Decompose the fraction into partial fractions: .
3. Solve for the constants A and B.
4. Integrate the resulting simpler fractions.
Step 3: Detailed Explanation or Calculation:
1. Partial Fraction Decomposition: To find A and B, we multiply by the common denominator: We can find the coefficients by substituting convenient values for .
Let : Let : So, the decomposition is: 2. Integration: Now we integrate the expression: Using the standard integral : Using the logarithm property : Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
For a vector function that describes a curve in space, the "slope" at a particular point is represented by the tangent vector to the curve at that point. The tangent vector is found by taking the first derivative of the vector function with respect to the parameter 't'.
Step 2: Key Formula or Approach:
To find the derivative of a vector function , we differentiate each component function with respect to 't': Step 3: Detailed Explanation:
The given vector function is: To find the tangent vector (slope), we differentiate with respect to t: Using the power rule for differentiation: Now, we need to find the tangent vector at the point t = 1. We substitute t = 1 into the derivative : Step 4: Final Answer:
The slope (tangent vector) at t = 1 is . So, option (A) is correct.

Step 1: Understanding the Concept:
To determine if a function is increasing or decreasing over an interval, we use its first derivative. A function is strictly increasing on an interval if its first derivative, , is positive ( ) for all in that interval.
Step 2: Key Formula or Approach:
1. Find the first derivative of the function, .
2. Analyze the sign of for all real numbers . We need to show that for all .
For a quadratic , if and the discriminant , the quadratic is always positive.
Step 3: Detailed Explanation:
The given function is: First, we find the derivative, : Now, we need to show that this quadratic expression is always positive for all . We can do this by checking its discriminant ( ).
Here, . Since the discriminant and the leading coefficient , the quadratic is always positive for all real values of .
Alternatively, we can complete the square: Since for all , the minimum value of is 0.
Therefore, the minimum value of is .
This means for all .
Step 4: Final Answer:
Since for all , the function is an increasing function in .

Step 1: Understanding the Concept:
The integral involves a rational function where the denominator is a product of linear factors. This suggests using the method of partial fraction decomposition to break down the integrand into simpler fractions that can be easily integrated.
Step 2: Key Formula or Approach:
We will decompose the fraction into the form: Multiplying both sides by , we get: To find the constants A and B, we can substitute the roots of the denominator.
Step 3: Detailed Explanation:
First, let's find the values of A and B.
Set : Next, set : Now, we can rewrite the integral as: Integrate each term separately: Using the standard integral : Using the property of logarithms : Step 4: Final Answer:
The evaluated integral is .

Step 1: Understanding the Concept:
This problem requires proving an identity involving inverse trigonometric functions. The key is to simplify the complex argument of the function using a suitable trigonometric substitution.
Step 2: Key Formula or Approach:
The presence of terms and suggests the substitution . This is because: We will also use the identity .
Step 3: Detailed Explanation:
Let the Left Hand Side (LHS) be .
Let . This implies , so .
Now, let's determine the range of . Given , we have .
The range of is . Therefore, . In this interval, both and are non-negative.
Now substitute into the expression: Substituting these into the LHS: Divide the numerator and denominator inside the parenthesis by (which is non-zero in the given range).
Using the identity , we can write: This is the formula for . Since , we have . This range is within the principal value branch of , i.e., .
So, we can simplify .
Substitute back .
Thus, LHS = RHS. Step 4: Final Answer:
The given identity is proved.

Step 1: Understanding the Concept:
A function is continuous at a point if:
1. is defined.
2. (the limit exists).
3. .

Step 2: Key Formula or Approach:
At , compute:
1. LHL:
2. RHL:
3.

Step 3: Detailed Calculation:
The function is defined as:


1. Left-Hand Limit (LHL):
.

2. Right-Hand Limit (RHL):
.

3. Value of the function:
.

Since LHL = 6 and RHL = -4, they are not equal. Hence, does not exist.

Step 4: Final Answer:
The function is discontinuous at .

Step 1: Understanding the Concept:
This integral is a classic example that can be solved efficiently using a special property of definite integrals, often called the "King's property". This property is particularly useful when the integrand has a certain symmetry with respect to the sum of the limits.
Step 2: Key Formula or Approach:
We will use the property of definite integrals: The strategy is to apply this property to the given integral, add the new form of the integral to the original one, and simplify the result.
Step 3: Detailed Explanation:
Let the given integral be I: Here, and . Their sum is .
Applying the property , we get: Using the trigonometric identity : Rewrite as : Now, add equation (1) and equation (2): Now, evaluate the simple integral: Finally, solve for I: Step 4: Final Answer:
We have shown that the value of the integral is . Hence proved.

Step 1: Understanding the Concept:
This is a standard form of a definite integral that can be solved using a property of definite integrals, often known as the "King's Rule", which helps to eliminate the 'x' in the numerator.
Step 2: Key Formula or Approach:
We use the property: .
Let the given integral be: Applying the property with : Since and , we have and . Step 3: Detailed Explanation:
Add equations (1) and (2): Let the integrand be . Since , we can use the property . Here . Divide the numerator and denominator by : Let , so . The limits of integration change from and . This is a standard integral of the form . Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
The problem describes the motion of a car. The key information is that the car "is stopped" at point Q. In physics and calculus, this means that the car's instantaneous velocity is zero at the time it reaches Q. Velocity is the first derivative of the distance function with respect to time.
Step 2: Key Formula or Approach:
1. The distance function is given: . 2. The velocity function is the derivative of the distance function: . 3. Set to find the time 't' when the car stops. 4. Substitute this value of 't' back into the distance function to find the total distance covered.
Step 3: Detailed Explanation:
First, expand the distance function for easier differentiation: Now, find the velocity function by differentiating with respect to t: The car stops at point Q, so its velocity is 0 at that time. We set : Factor out the common term : This gives two possible solutions for t: corresponds to the starting point P, where the car was at rest. The other solution gives the time to reach point Q. So, the time required for the car to reach point Q is 4 seconds.
Now, to find the distance between P and Q, we substitute into the distance function : The distance between P and Q is 16 metres.
Step 4: Final Answer:
The time required to reach point Q is 4 seconds, and the distance between P and Q is 16 metres.

Step 1: Understanding the Concept:
This is an indefinite integral that can be solved using the method of substitution (u-substitution). We look for a part of the integrand whose derivative is also present (up to a constant factor).
Step 2: Key Formula or Approach:
1. Identify a suitable substitution, . Let .
2. Differentiate with respect to to find .
3. Rewrite the integral entirely in terms of and .
4. Integrate with respect to .
5. Substitute the original expression for back into the result.
Step 3: Detailed Explanation or Calculation:
Let the integral be .
1. Substitution:
Notice that the derivative of the exponent is , which is a constant multiple of the other factor in the integrand. So, we choose: Let .
2. Differentiate:
Then, .
This implies .
3. Rewrite the integral:
Substitute and into the integral: 4. Integrate:
The integral of is . 5. Substitute back:
Replace with : Step 4: Final Answer:
The value of the integral is , where C is the constant of integration.

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). We need to find the minimum value of a linear objective function within a feasible region defined by a set of linear inequalities. The minimum value, if it exists, will occur at one of the corner points (vertices) of the feasible region.
Step 2: Identifying the Feasible Region and Corner Points:
First, we graph the lines corresponding to each constraint:



The feasible region is the area in the first quadrant ( ) that satisfies all inequalities:
(Below )
(Above )
(Above )
The vertices of the feasible region are the points of intersection of these boundary lines. Point A: Intersection of and -axis ( ).
. So, A = (0, 3). Point B: Intersection of and -axis ( ).
. So, B = (1, 0). Point C: Intersection of and -axis ( ).
. So, C = (6, 0).
The feasible region is unbounded in the first quadrant.
Step 3: Evaluating the Objective Function at Corner Points:
We evaluate the objective function at each corner point.
At A(0, 3):
At B(1, 0):
At C(6, 0):
Step 4: Final Answer:
The minimum value of the objective function at the corner points of the feasible region is -300, which occurs at the point (6, 0).

Step 1: Understanding the Concept:
This is an optimization problem where we need to find the maximum volume of an open-top box. The volume is expressed as a function of the side 'x' of the square cutouts, and calculus is used to find the maximum value of this function.
Step 2: Key Formula or Approach:
1. Express the dimensions (length, width, height) of the box in terms of x.
2. Formulate the volume function, .
3. Find the derivative and set it to zero to find critical points.
4. Use the second derivative test, , to confirm that the critical point corresponds to a maximum.
Step 3: Detailed Explanation:
The original sheet has dimensions L = 16 m and W = 6 m.
When squares of side x are cut from each corner, the dimensions of the folded box become:
Height:
Length:
Width:
The volume is given by : For the dimensions to be physically possible, , , and . The valid domain for x is .
Find the first derivative to locate critical points: Set : Divide by 4: Using the quadratic formula, : This gives two possible values: and .
Within our domain , the only valid critical point is .
Use the second derivative test to confirm it's a maximum: Since , the volume is maximized at .
Calculate the maximum volume: Step 4: Final Answer:
The maximum volume of the box is cubic meters.

Step 1: Understanding the Concept:
We can prove this identity by converting one inverse trigonometric function to another. A common method is to use a right-angled triangle to represent the relationship given by the first function and then derive the second function from the triangle's dimensions.
Step 2: Key Formula or Approach:
1. Let . This implies .
2. Construct a right-angled triangle where .
3. Use the Pythagorean theorem to find the length of the adjacent side.
4. From the triangle, find and then express in terms of .
Step 3: Detailed Explanation:
Let . Then, by definition, . We can write this as .
Consider a right-angled triangle with angle . We can set:
Opposite side =
Hypotenuse =
Using the Pythagorean theorem ( ): Now, we can find the value of from this triangle: Taking the inverse tangent of both sides, we get: Since we started with , we have proven that: Step 4: Final Answer:
The identity is proven by equating the two expressions for .

Step 1: Understanding the Concept:
This question requires evaluating the indefinite integral of and then comparing the result with the given expression to find the function .
Step 2: Key Formula or Approach:
We will use the method of integration by parts to solve . The formula for integration by parts is: Step 3: Detailed Explanation:
To evaluate , we can write it as . Let's choose our u and dv according to the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). Let and .
Then, we find du and v: Now, applying the integration by parts formula: Now, we compare this result with the given equation: By direct comparison of the terms, we can see that: Step 4: Final Answer:
The function is . Therefore, the correct option is (iii).

Step 1: Understanding the Concept:
The area enclosed between two curves is found by integrating the difference between the upper function and the lower function over the interval defined by their points of intersection.
Step 2: Key Formula or Approach:
1. Find the points of intersection by setting the two equations equal to each other.
2. Set up the definite integral: Area , where and are the x-coordinates of the intersection points.
3. Evaluate the integral.
Step 3: Detailed Explanation:
The given curves are (a parabola) and (a horizontal line).
To find the points of intersection, we set . This gives . The points of intersection are (-4, 16) and (4, 16).
In the interval [-4, 4], the line is above the parabola . So, and .
The area is given by the integral: Since the integrand is an even function and the limits of integration are symmetric about 0, we can simplify the calculation: Now, we evaluate the integral: Step 4: Final Answer:
The area enclosed by the curve and the line is square units.

Step 1: Understanding the Concept:
The problem asks to find the derivative of with respect to from an implicit equation. The strategy is to first simplify the equation algebraically to express explicitly as a function of , and then differentiate.
Step 2: Key Formula or Approach:
The main approach involves algebraic manipulation to isolate , followed by differentiation using the quotient rule.
Quotient Rule: If , then .
Step 3: Detailed Explanation or Calculation:
Given the equation: Rearrange the terms: Square both sides to eliminate the square roots: Expand the terms: Rearrange the equation to group terms: Factorize the terms. The first two terms form a difference of squares, and we can factor from the last two terms: Factor out the common term : This implies either or .
If , then . Substituting this into the original equation gives , which is only true for or . Since we are looking for a general relation for , we consider the other case.
So, we use the relation: Now, we express as a function of : Now, we differentiate with respect to using the quotient rule: Step 4: Final Answer:
We have successfully shown that if , then .

Step 1: Understanding the Concept:
We need to evaluate a definite integral of over a symmetric interval . We can use properties of even and odd functions to simplify the integral. Then, we use a trigonometric identity to integrate .
Step 2: Key Formula or Approach:
1. Property of definite integrals over symmetric intervals: If is an even function ( ), then .
2. Power-reducing identity: .
Step 3: Detailed Explanation or Calculation:
Let . Let's check if it is an even function: Since is an even function, we can simplify the integral: Now, use the power-reducing identity: Integrate term by term: Evaluate at the limits: Since : Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
We need to find the indefinite integral of . Since there is no direct standard formula for integrating , we must first rewrite it using a trigonometric identity to reduce the power.
Step 2: Key Formula or Approach:
The half-angle identity (or power-reducing formula) for cosine is:
We will substitute this into the integral and then integrate term by term.
Step 3: Detailed Explanation or Calculation:
Split the integral into two parts:
To integrate , we get .
Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
This definite integral can be solved using the properties of definite integrals. The presence of 'x' in the numerator suggests using the property , which often helps to eliminate the 'x' term.
Step 2: Key Formula or Approach:
1. Use the property .
2. Add the two forms of the integral to eliminate x from the numerator.
3. Use the property if .
4. Evaluate the resulting integral by dividing the numerator and denominator by and using a substitution.
Step 3: Detailed Explanation or Calculation:
Let the given integral be . Using the property :
Since and , their squares are and . Adding equations (1) and (2):
Let . Since , we can use the property . Here . Divide numerator and denominator by : Let , so . The limits change from and . This is a standard integral form: Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
This definite integral can be evaluated using the properties of definite integrals and trigonometric identities. The "King's property" is a key tool for such problems.
Step 2: Key Formula or Approach:
1. Let .
2. Apply the property with .
3. Add the two expressions for I.
4. Use trigonometric and logarithm properties to simplify the resulting integral.
5. Use the property if .
Step 3: Detailed Explanation or Calculation:
Let
Using the property :
Since :
Adding equations (1) and (2):
Using the logarithm property :
Using the identity :
Let's check if the integrand is symmetric about . .
So we can use the property . There is a standard result for this integral: .
Let's prove this result. Let . Using , . Adding these two forms of : Let . Limits: . So, . This gives .
Now substitute this result back into our main problem: Step 4: Final Answer:
Hence, it is proved that .

Step 1: Understanding the Concept:
The parabola opens to the right with its vertex at the origin (0, 0). Its latus rectum is a line segment perpendicular to the axis of symmetry (the x-axis), passing through the focus at . The equation of the line containing the latus rectum is therefore . The area is bounded by the parabola and this line.
Step 2: Key Formula or Approach:
The area of a region bounded by a curve , the x-axis, and the lines and is given by .
Since the parabola is symmetric about the x-axis, we can find the area in the first quadrant (from to ) and multiply it by 2.
Step 3: Detailed Explanation or Calculation:
From , we get .
For the area in the first quadrant, we take .
The limits of integration are from the vertex to the latus rectum .
The area in the first quadrant is:
The total area is twice the area in the first quadrant:
Step 4: Final Answer:
The area of the region is square units.

Step 1: Understanding the Concept:
To find the equation of a tangent line to a curve at a given point, we first need to find the slope of the tangent, which is the value of the derivative at that point. The slope of the normal line is the negative reciprocal of the tangent's slope.
Step 2: Key Formula or Approach:
1. Differentiate the curve's equation implicitly with respect to x to find .
2. Evaluate at the point (1, 1) to get the slope of the tangent, .
3. Calculate the slope of the normal: .
4. Use the point-slope form of a line, , to find the equations.
Step 3: Detailed Explanation or Calculation:
The equation of the curve is .
Differentiating both sides with respect to x:
Divide by :
Solve for :
Now, find the slope of the tangent at the point (1, 1):
The slope of the normal is:
Equation of the Tangent:
Using the point-slope form with point (1, 1) and slope :
Equation of the Normal:
Using the point-slope form with point (1, 1) and slope :
Step 4: Final Answer:
The equation of the tangent is .
The equation of the normal is .

Step 1: Understanding the Concept:
This is a definite integral problem. We need to find the antiderivative of the integrand and then evaluate it at the upper and lower limits of integration, according to the Fundamental Theorem of Calculus.
Step 2: Key Formula or Approach:
The standard integral formula required is:
The Fundamental Theorem of Calculus states:
, where .
Step 3: Detailed Explanation or Calculation:
First, find the antiderivative of the integrand:
Now, apply the limits of integration from 1 to :
We need to find the principal values for these inverse trigonometric functions.
- The angle whose tangent is is . So, .
- The angle whose tangent is 1 is . So, .
Now, perform the subtraction:
Step 4: Final Answer:
The value of the definite integral is .

Step 1: Understanding the Concept:
This is a standard problem in definite integrals that can be solved efficiently using a property of definite integrals, often known as the "King's Rule".
Step 2: Key Formula or Approach:
The property states that for a continuous function on : Step 3: Detailed Calculation:
Let the given integral be . First, express in terms of and : Now, apply the property with . Using the trigonometric identities and , we get: Now, add equation (1) and equation (2): Step 4: Final Answer:
The value of the integral is .

Step 1: Understanding the Concept:
This is a problem of evaluating a definite integral. A common and powerful technique for integrals of the form is to use the property . This property is sometimes called the "King Property".
Step 2: Key Formula or Approach:
1. Let the given integral be .
2. Apply the property , where .
3. Use the trigonometric identity .
4. Simplify the resulting integral and add it to the original integral to solve for .
Step 3: Detailed Explanation or Calculation:
Let the integral be : Using the property , with : We know the identity .
Substituting this into the integral: Simplify the term inside the logarithm: Using the logarithm property : Split the integral into two parts: From equation (1), we know that the second term is . So, we have: Since is a constant: Solving for : Step 4: Final Answer:
We have successfully proven that .

Step 1: Understanding the Concept:
This problem involves solving an indefinite integral containing trigonometric functions. The key to solving it is to simplify the integrand using trigonometric identities before performing the integration, which will likely involve a substitution.
Step 2: Key Formula or Approach:
The solution uses the following key trigonometric identities: 1. and 2. 3. 4. 5. The integration will be performed using the method of substitution.
Step 3: Detailed Explanation or Calculation:
Let the integral be . First, simplify the denominator: Using the double angle identities, and : Now substitute this back into the integral: Rewrite the integrand in terms of and . Since : Now, we use the substitution method. Let . Then, the derivative is .
This gives us .
Substitute and into the integral: Now, integrate with respect to : Finally, substitute back : Step 4: Final Answer:
The solution to the integral is .

Step 1: Understanding the Concept:
The problem is to find the total area enclosed by an ellipse. The equation of an ellipse is given in standard form, from which we can identify the lengths of the semi-major and semi-minor axes.
Step 2: Key Formula or Approach:
The standard equation of an ellipse centered at the origin is: where is the length of the semi-horizontal axis and is the length of the semi-vertical axis. The area of such an ellipse is given by the formula: Step 3: Detailed Explanation or Calculation:
The given equation of the ellipse is: Comparing this to the standard form , we can identify and : Now, we use the formula for the area of an ellipse: Substitute the values of and : Step 4: Final Answer:
The area bounded by the given ellipse is square units.

Step 1: Understanding the Concept:
This is an optimization problem that falls under the application of derivatives. We need to find the dimensions of an inscribed cylinder that maximize its curved surface area. The key is to express the quantity to be maximized (curved surface area) as a function of a single variable and then use calculus to find the maximum.
Step 2: Key Formula or Approach:
1. Define the dimensions of the cone (radius , height ) and the inscribed cylinder (radius , height ).
2. Use similar triangles from a cross-sectional view to establish a relationship between , , , and .
3. Write the formula for the curved surface area (CSA) of the cylinder, .
4. Use the relationship from step 2 to express as a function of a single variable, say .
5. Find the first derivative of with respect to ( ) and set it to zero to find critical points.
6. Use the second derivative test ( ) to confirm that the critical point corresponds to a maximum.
Step 3: Detailed Explanation or Calculation:
Let the cone have a fixed radius and height . Let the cylinder inscribed within it have a variable radius and height .
Consider a vertical cross-section of the cone and cylinder. This forms two similar right-angled triangles. The larger triangle has height and base . The smaller triangle, which is above the cylinder, has height and base .
By the property of similar triangles: We can express in terms of (and the constants ): The curved surface area (CSA) of the cylinder is given by .
Substitute the expression for into the formula for to get as a function of : To find the maximum area, we differentiate with respect to and set the derivative to zero: Set for critical points: Since , we must have: To confirm this is a maximum, we use the second derivative test: Since and are positive dimensions, is negative. A negative second derivative indicates that the function has a local maximum at the critical point.
Step 4: Final Answer:
The curved surface area of the inscribed cylinder is maximum when its radius is equal to , which is half the radius of the cone. This completes the proof.

Step 1: Understanding the Concept:
This is an indefinite integral of a rational function. The standard method for solving such integrals is to use partial fraction decomposition, which involves breaking down the complex fraction into a sum of simpler fractions that are easier to integrate.
Step 2: Key Formula or Approach:
1. Factor the denominator completely.
2. Express the rational function as a sum of partial fractions. For a denominator of the form , the decomposition is .
3. Solve for the unknown coefficients (A, B, C).
4. Integrate the resulting simpler fractions using standard integration formulas: and .
Step 3: Detailed Explanation or Calculation:
1. Factor the denominator: Factor by grouping: So the integral is: 2. Decompose into partial fractions: Multiply both sides by the denominator : 3. Solve for A, B, and C:
We can substitute convenient values of to find the coefficients.
Let : Let : To find A, let's use another value, for example : Substitute the values of B and C we found: 4. Integrate:
Now, substitute the coefficients back into the integral: Integrate term by term: Step 4: Final Answer:
The solution to the integral is: This can also be written using logarithm properties as: