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.

Let us consider $$I=\int \frac{dx}{x\sqrt{1+x^2}}$$ and make $u=\sqrt{1+x^2}$ that is to say $x=\sqrt{u^2-1}$ and then $dx=\frac{u}{\sqrt{u^2-1}}du$. So, $$I=\int \frac{du}{u^2-1}$$ Now, use partial fraction decomposition and you are done.
