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