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

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

The correct answer is Option (2) → $\frac{1}{2} \tan^{-1} \left( \frac{x - 3}{2} \right) + C$

We have $x^2 - 6x + 13 = x^2 - 6x + 3^2 - 3^2 + 13 = (x - 3)^2 + 4$

So, $\int \frac{dx}{x^2 - 6x + 13} = \int \frac{dx}{(x - 3)^2 + 2^2}$

Let $x - 3 = t$. Then $dx = dt$

Therefore, $\int \frac{dx}{x^2 - 6x + 13} = \int \frac{dt}{t^2 + 2^2} = \frac{1}{2} \tan^{-1} \frac{t}{2} + C$

$= \frac{1}{2} \tan^{-1} \frac{x - 3}{2} + C$