Answer
$$\frac{1}{54}\left(\tan ^{-1} \frac{x}{3}+\frac{x}{\sqrt{x^{2}+9}} \cdot \frac{3}{\sqrt{x^{2}+9}}\right)+C$$
Work Step by Step
Given $$\int \frac{d x}{\left(x^{2}+9\right)^{2}}$$
Since
\begin{align*}
x&=3\tan u\ \ \\
dx&=3\sec^2 udu
\end{align*}
Then
\begin{align*}
\int \frac{d x}{\left(x^{2}+9\right)^{2}}&=\int \frac{3\sec^2 udu}{\left(9\tan^2 u+9\right)^{2}}\\
&= \frac{1}{27}\int \frac{du}{\sec^2 u}\\
&=\frac{1}{27}\int\cos^2 udu\\
&= \frac{1}{54} \int (1+\cos 2u)du\\
&= \frac{1}{54} \left(u+ \frac{1}{2}\sin 2u\right)+C\\
&= \frac{1}{54} \left(u+ \sin u\cos u\right)+C\\
&= \frac{1}{54}\left(\tan ^{-1} \frac{x}{3}+\frac{x}{\sqrt{x^{2}+9}} \cdot \frac{3}{\sqrt{x^{2}+9}}\right)+C
\end{align*}
