Answer
$$\frac{1}{2}\tan^{-1}\frac{x+1}{2}+C$$
Work Step by Step
Given $$\int \frac{d x}{x^{2}+2 x+5}$$
\begin{aligned}
\int \frac{1}{x^{2}+2 x+5} d x &=\int \frac{1}{(x+1)^{2}+4} d x
\end{aligned}
Let $$x+1=2\tan u,\ \ \ \ \ \ \ \ dx=2\sec^2 udu $$
Then
\begin{align*}
\int \frac{1}{x^{2}+2 x+5} d x &=\int \frac{2\sec^2 udu }{4\tan^{2}u+4} \\
&=\int \frac{2\sec^2 udu }{4\sec^{2}u } \\
&=\frac{1}{2}\int du\\
&=\frac{1}{2}u+C\\
&=\frac{1}{2}\tan^{-1}\frac{x+1}{2}+C\\
\end{align*}
