Answer
$$\frac{1}{3}\tan^{-1}\left(\frac{x+4}{3}\right)+C $$
Work Step by Step
Given $$ \int \frac{d x}{x^{2}+8 x+25}$$
Since
$$x^2+8x+25= (x+4)^2+9 $$
Then
\begin{aligned}
\int \frac{d x}{x^{2}+8 x+25} &=\int \frac{d x}{(x+4)^{2}+3^{2}}\\
&=\frac{1}{3}\tan^{-1}\left(\frac{x+4}{3}\right)+C
\end{aligned}
