<s>[INST] 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.
 [/INST] Maybe another substitution is interesting:
For $x>0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = \int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = -\int\frac{dv}{\sqrt{v^2+1}} = - \ln(v+\sqrt{v^2+1})=$$
$$ =- \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
For $x<0, v=\frac{1}{x}$
$$
\int \frac 1 {x\sqrt{1+x^2}} \,dx = -\int \frac {1} {x^2\sqrt{\frac{1}{x^2}+1}} \,dx = \int\frac{dv}{\sqrt{v^2+1}} =  \ln(v+\sqrt{v^2+1})+ C=$$
$$ = \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right) + C.
$$
</s>