Answer
$$\frac{1}{48} \tan ^{-1}\left(\frac{3 t}{2}\right)+\frac{t}{8\left(4+9 t^{2}\right)}+C$$
Work Step by Step
Given $$\int \frac{d t}{\left(9 t^{2}+4\right)^{2}}$$
Let
$$3t=2\tan u\ \ \ \ \ \to\ \ \ \ 3dt= 2\sec^2 udu $$
Then
\begin{align*}
\int \frac{d t}{\left(9 t^{2}+4\right)^{2}}&=\frac{2}{3} \int \frac{\sec^2 udu }{\left(4\tan ^2 u+4\right)^{2}}\\
&= \frac{1}{24} \int \frac{\sec^2 udu }{\sec^{4}u}\\
&= \frac{1}{24}\int \cos^2 udu\\
&= \frac{1}{48}\int (1+\cos 2u)du\\
&=\frac{1}{48}\left[u+\frac{\sin 2 u}{2}\right]+C\\
&=\frac{1}{48} u+\frac{1}{48} \sin u \cos u+C\\
& = \frac{1}{48} u+\frac{1}{48} \sin u \cos u+C \\
&=\frac{1}{48} \tan ^{-1}\left(\frac{3 t}{2}\right)+\frac{1}{48} \frac{3 t}{\sqrt{4+9 t^{2}}} \frac{2}{\sqrt{4+9 t^{2}}}+C\\
&=\frac{1}{48} \tan ^{-1}\left(\frac{3 t}{2}\right)+\frac{t}{8\left(4+9 t^{2}\right)}+C
\end{align*}
