Answer
$\int\frac{1}{x^2\sqrt{1+x^2}}dx=-\frac{\sqrt{1+x^2}}{x}+C$
Work Step by Step
Verify $\int\frac{1}{x^2\sqrt{1+x^2}}dx=-\frac{\sqrt{1+x^2}}{x}+C$
We take the derivative of the right side of this equation and verify that it equals the inside of our integral.
\begin{equation*}
\frac{d}{dx}\left(-\frac{\sqrt{1+x^2}}{x}+C\right)=\frac{d}{dx}\left(-\frac{\sqrt{1+x^2}}{x}\right)+\frac{d}{dx}C=\frac{d}{dx}\left(-\frac{\sqrt{1+x^2}}{x}\right)
\end{equation*}
\begin{equation*}
=-\frac{x\times\frac{1}{2}(1+x^2)^{-\frac{1}{2}}\times 2x-\sqrt{1+x^2}}{x^2}
\end{equation*}
\begin{equation*}
=-\frac{\frac{x^2}{\sqrt{1+x^2}}-\sqrt{1+x^2}}{x^2}=-\frac{\frac{x^2-(1+x^2)}{\sqrt{1+x^2}}}{x^2}=-\frac{\frac{-1}{\sqrt{1+x^2}}}{x^2}=\frac{1}{x^2\sqrt{1+x^2}}
\end{equation*}
Hence, the result is proved.
