How to integrate $\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$ I want to know how to integrate 
$$\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx$$
Could anyone solve it?
Thanks

$$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{x\sqrt{1+x^2}}{x^2}\,dx$$
Let $u=x^2+1, du =2x dx$. Then
$$\int\frac{\sqrt{1+x^2}}{x}\,dx= \frac{1}{2}\int\frac{\sqrt{u}}{u^2-1}\,du$$
If $v=\sqrt{u}$ the n $u=v^2, du=2vdv$. Thus
$$\int\frac{\sqrt{1+x^2}}{x}\,dx=\int\frac{v^2}{v^4-1}\,dv$$
By Partial Fraction Decomposition 
$$\frac{v^2}{v^4-1}=\frac{A}{v-1}+\frac{B}{v+1}+\frac{Cv+D}{v^2+1}$$
find $A,B,C,D$, and use $v=\sqrt{u}=\sqrt{x^2+1}$ and you are done.
