$\int \frac{1}{x^2+4 x+13} d x$ is equal to

$\log \left(x^2+4 x+13\right)+C$ $\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)$ $\log (2 x+4)+C$ $\frac{2 x+4}{\left(x^2+4 x+13\right)^3}+C$

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

We have, $I =\int \frac{1}{x^2+4 x+13} d x$ $=\int \frac{1}{(x+2)^2+3^2} d x=\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)+C$