Find: $\int \frac{dx}{x^2 - 6x + 13}$

$\frac{1}{2} \tan^{-1}\left(\frac{x - 3}{2}\right) + C$

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 $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$]