Find: $\int \frac{dx}{x^2 - 6x + 13}$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{2} \tan^{-1}\left(\frac{x - 3}{2}\right) + C$
Given integral is:
$I = \int \frac{dx}{x^2 - 6x + 13}$
$= \int \frac{dx}{(x - 3)^2 + 13 - 9}$
$= \int \frac{dx}{(x - 3)^2 + 4}$
$= \int \frac{dx}{(x - 3)^2 + 2^2}$
$= \frac{1}{2} \tan^{-1} \left( \frac{x - 3}{2} \right) + C$
[Using $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$]