Integral

Detailed Solution:

Step 1: Substitution
Let t = eˣ ⇒ dt = eˣ dx

Given integral becomes:
I = ∫ [t / ((2 + t)(t + 1))] × (1 / t) dt

This simplifies to:
I = ∫ [1 / ((2 + t)(1 + t))] dt

Step 2: Partial Fraction Decomposition
Break the expression into partial fractions:

1 / ((t + 1)(t + 2)) = A / (t + 1) + B / (t + 2)

Multiplying both sides by (t + 1)(t + 2):
1 = A(t + 2) + B(t + 1)

Now solve for A and B:

• Put t = -2 ⇒ 1 = A(0) + B(-1) ⇒ B = -1
• Put t = -1 ⇒ 1 = A(1) + B(0) ⇒ A = 1

So, the integrand becomes:
I = ∫ [1 / (t + 1) - 1 / (t + 2)] dt

Step 3: Integrate

I = ∫ [1 / (t + 1)] dt - ∫ [1 / (t + 2)] dt
I = log|t + 1| - log|t + 2| + C

Using the logarithmic identity:
log A - log B = log (A / B)

I = log((t + 1) / (t + 2)) + C

Step 4: Substitute back t = eˣ

I = log((eˣ + 1) / (eˣ + 2)) + C

Final Answer:
log((eˣ + 1) / (eˣ + 2)) + C

Correct Option: (D)

We have f(x) = sin|x| + cos|x|
Then, f(x) =f(–x) Since, (f(x) is an even function.
I =
I = 2
I = 2[-cosx + sinx]^π/2₀
I = 2[-cos + sin + cos 0 - sin 0]
I = 2[0 + 1 + 1–0]
I = 2x2
I = 4
Therefore, the correct option is (C) 4

The given differential equation is:

Step 1: Simplify the equation. To identify the degree of the equation, first eliminate the square root by squaring both sides. Rearranging the terms gives: Squaring both sides results in: This simplifies to:

Step 2: Determine the order and degree.

The order of a differential equation is defined as the highest derivative present in the equation. Here, the highest derivative is , so the order is .
The degree refers to the power of the highest order derivative after removing any radicals or fractional powers. Upon squaring, the highest power of is .

Final Answer:

Step 1: Simplify the integral The given integral is: We can simplify this by separating terms:

Step 2: Solve each integral The second integral can be solved easily by substitution, let: This gives: Now, the first integral can be simplified using substitution as well: This can be solved by standard methods for integrating square roots.

Answer: The final result is:

We are given the integral:

Step 1: Complete the square

The expression inside the square root is . Completing the square:

Thus, the integral becomes:

Step 2: Substitution

We perform the substitution , which gives . The integral becomes:

Step 3: Use the standard formula for this type of integral

We use the standard integral formula for , which is:

Substituting into this formula, we get:

Step 4: Back-substitute

Now, substitute back into the result:

Conclusion:

The final solution is: