Answer
$$ \ln|\sqrt{1+x^2}+x| -\frac{\sqrt{x^{2}+1}}{x}+C$$
Work Step by Step
Given
$$ \int \frac{\sqrt{x^{2}+1}}{x^{2}} d x$$
Let
$$x=\tan u\ \ \ \ \ \ \ \ \ \ \ \ dx=\sec^2 udu $$
Then
\begin{align*}
\int \frac{\sqrt{x^{2}+1}}{x^{2}} d x&=\int \frac{\sqrt{\tan^{2}u+1}}{\tan^{2}u} \sec^2 udu\\
&= \int \frac{\sqrt{\sec^{2}u}}{\tan^{2}u}(1+ \tan^2 u)du\\
&= \int (\sec u +\cot u\csc u)du\\
&=\ln|\sec u+\tan u| -\csc u+C\\
&= \ln|\sqrt{1+x^2}+x| -\frac{\sqrt{x^{2}+1}}{x}+C
\end{align*}
