How can we find $\int\frac{1}{x^2\sqrt{x^2+1}}dx$? How can we integrate 
$$\int\frac{1}{x^2\sqrt{x^2+1}}dx$$
I tried some substitutions but I failed . I tried for example $$v=\sqrt{1+x^2}$$
 A: 
Another way to evaluate the integral, without making a trigonometric or hyperbolic substitution, is to make the substitution $x=1/t$.  

Proceeding reveals 
$$\begin{align}
\int \frac{1}{x^2\sqrt{1+x^2}}\,dx&\overbrace{=}^{x=1/t}-\int \frac{t}{\sqrt{t^2+1}}\,dt\\\\
&=-\sqrt{1+t^2}+C\\\\
&\overbrace{=}^{t=1/x}-\frac{\sqrt{1+x^2}}{x}+C
\end{align}$$
And we are done!
A: Taking $x=\tan t\implies \mathrm dx=\sec^2t \mathrm dt$ gives the integral 
$$
\int \frac{\cos t}{\sin^2 t}\mathrm d t
$$
where you can take $u=\sin t$ to finish, 
$$
\int \frac{1}{x^2\sqrt{1+x^2}}\mathrm dx=\int \frac{\cos t}{\sin^2 t}\mathrm d t
=\int u^{-2}\mathrm du\\=-u^{-1}+c=-\frac{1}{\sin t}+C\\
=-\frac{1}{\sin\arctan x}+C\\
=-\frac{\sqrt{1+x^2}}{x}+C
$$
A: You can use $x=\sinh(t)$.
Then $\mathrm dx=\cosh (t) \mathrm dt$.
The integral using the substitution:
$$
\int \frac{1}{x^2\sqrt{1+x^2}}\mathrm dx=\int \frac{\cosh (t) }{\sinh^2(t)\sqrt{1+\sinh^2(t)}}\mathrm d t
=\int \frac{1}{\sinh^2(t)}\mathrm dt\\=-\coth(t)+C=-\coth(\sinh^{-1}(x))+C
=-\frac{\sqrt{1+x^2}}{x}+C
$$
