Integration By Partial Fractions

To find the value of , we must first find a common denominator for the fractions. The denominators are 6 and 4. The least common multiple (LCM) of 6 and 4 is 12. We will convert both fractions to have this common denominator.

can be rewritten as .

can be rewritten as .

Now, we add the two fractions:

Thus, the correct value is .

Given :
Multiplying both sides by , we get: Expanding the RHS: Combining like terms: Now, equate coefficients with LHS : Solving equations: From (1):
Substitute into (2): From (3): Add (4) and (3):
So, , and Thus, Integrating: Break into parts: Compare with: So,

We are given: Now split the integral: The first term: The second term: This integral can be solved using partial fractions. The result is: So, the combined result is:

Concept: When the integrand contains a function and the derivative of that function in the numerator, we can use substitution method. Observe the expression: Its derivative is which closely resembles the numerator.
Step 1: {Choose substitution.} Let Then
Step 2: {Rewrite the integrand.} Using the substitution and simplifying the expression, the integral transforms into
Step 3: {Integrate the expression.}
Step 4: {Substitute back the value of .} Therefore, Hence, the correct answer is (B).

Concept: For rational functions where the degree of the numerator is less than the denominator, we use partial fraction decomposition. If is given, it can be written as Integrals of the form are then applied.
Step 1: {Factor the denominator.} Thus,
Step 2: {Apply partial fractions.} Assume Multiplying by : Comparing coefficients: Solving,
Step 3: {Substitute back into the integral.}
Step 4: {Integrate.}