Answer
$$\frac{1}{250} \tan ^{-1}\left(\frac{x}{5}\right)+\frac{x}{50\left(x^{2}+25\right)}+C$$
Work Step by Step
Given $$\int \frac{d x}{\left(x^{2}+25\right)^{2}} $$
Let $$x=5\tan u\ \ \ \ \ \ \ \ \ \ \ \ dx=5\sec^2 udu $$
Then
\begin{align*}
\int \frac{d x}{\left(x^{2}+25\right)^{2}}&=\int \frac{5\sec^2 udu}{\left(25\tan^{2}u+25\right)^{2}}\\
&=\int \frac{5\sec^2 udu}{\left(25\sec^{2}u\right)^{2}}\\
&= \frac{1}{125}\int \frac{du}{\sec^2 u}\\
&= \frac{1}{125}\int \cos^2 udu\\
&= \frac{1}{250}\int (1+\cos2 u)du\\
&=\frac{1}{250}(u+\frac{1}{2}\sin 2u)+C\\
&=\frac{1}{250}(u+ \sin u\cos u)+C\\
&=\frac{1}{250} \tan ^{-1} \frac{x}{5}+\frac{1}{250} \cdot \frac{5}{\sqrt{x^{2}+25}} \cdot \frac{x}{\sqrt{x^{2}+25}}+C\\
&=\frac{1}{250} \tan ^{-1}\left(\frac{x}{5}\right)+\frac{x}{50\left(x^{2}+25\right)}+C
\end{align*}
