Integral of $\frac{\sqrt{x^2+1}}{x}$ So I have to do the integration of $\frac{\sqrt{x^2+1}}{x}$. Give me a hint. What should I replace? Should I do it with integration by parts?

You could use the substitution
$$
\begin{align}
&&u&=x+\sqrt{x^2+1},&x&=\frac{u^2-1}{2u} \\
&&\sqrt{x^2+1}&=\frac{u^2+1}{2u},&dx&=\frac{u^2+1}{2u^2}du
\end{align}
$$
to get (unless I did some mistake)
$$
\int\frac{\sqrt{x^2+1}}{x}dx=
\int\frac{\frac{u^2+1}{2u}}{\frac{u^2-1}{2u}}\frac{u^2+1}{2u^2}du=
\int\frac{(u^2+1)^2}{2u^2(u^2-1)}du=
\int\frac{(u^2-1)^2+4u^2}{2u^2(u^2-1)}du=
$$
$$
=\int\left(\frac{(u^2-1)^2}{2u^2(u^2-1)}+\frac{4u^2}{2u^2(u^2-1)}\right)\,du=
\int\left(\frac{u^2-1}{2u^2}+\frac{2}{u^2-1}\right)\,du=
$$
$$
=\frac{u}{2}+\frac{1}{2u}+\ln|u-1|-\ln|u+1|+C.
$$
Now just do the backsubstitution.
