Answer
$$ \frac{ t}{\sqrt{1+4 t^{2}}}+C$$
Work Step by Step
Given $$\int \frac{d t}{\left(1+4 t^{2}\right)^{3 / 2}}$$
Let $$ 2t=\tan u\ \ \ \ \ \ 2dt =\sec^2 udu$$
Then
\begin{aligned}
\int \frac{1}{\left(1+4 t^{2}\right)^{3 / 2}} d t
&=\int \frac{1}{\left(1+\tan ^{2} u\right)^{3 / 2}} \frac{\sec ^{2} u d u}{2} \\
&=\int \frac{1}{\left(\sec ^{2} u\right)^{3 / 2}} \frac{\sec ^{2} u d u}{2} \\
&=\int \frac{1}{\sec ^{3} u} \frac{\sec ^{2} u d u}{2} \\
&=\int \frac{d u}{2 \sec u} \\
&=\frac{1}{2} \int \cos u \, d u \\
&=\frac{1}{2} \sin u+C \\
&= \frac{ t}{\sqrt{1+4 t^{2}}}+C
\end{aligned}
