Question

$\int \frac{dx}{x^2 + 4x + 13}$ is equal to

• A. $\log(x^2 + 4x + 13) + c$
• B. $\frac{1}{3} \tan^{-1} \left( \frac{x+2}{3} \right) + c$
• C. $\log(2x+4) + c$
• D. $\frac{1}{x^2 + 4x + 13} + c$

Hint:

brush up your derivatives concept and inverse formulas

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Complete the square in the denominator.
Step 2: Detailed Explanation:
$x^2 + 4x + 13 = (x+2)^2 + 9 = (x+2)^2 + 3^2$.
So $\int \frac{dx}{(x+2)^2 + 3^2} = \frac{1}{3} \tan^{-1} \left( \frac{x+2}{3} \right) + c$.
Step 3: Final Answer:
The integral is $\frac{1}{3} \tan^{-1} \left( \frac{x+2}{3} \right) + c$.