How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution? How you integrate 
$$\frac{1}{x\sqrt{1+x^2}}$$
using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$ 
And now I don't know how to proceed using substitution rule.

If $u=\sqrt{1+x^2}$ then $u^2 = 1+x^2$, so $x^2= u^2-1$.  Then you have
\begin{align}
& \int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {x} {x^2\sqrt{1+x^2}} \,dx \\[8pt]
= {} & \int\frac{du}{u^2-1} = \int\frac{du}{(u-1)(u+1)}.
\end{align}
Then use partial fractions.
