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?
 A: Hint: substitute $x = \tan{y}$; $dx = \sec^2{y} dy$.  Then $x^2+1 = \sec^2{y}$.
A: 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.
A: This problem is a lot uglier than it originally looked.  First, let's bring the denominator under the radical.
$$\int\sqrt{\frac{x^2+1}{x^2}}dx=\int\sqrt{1+\frac1{x^2}}dx$$.
Now let's make the substitution rlgordonna suggested.
$$x=\tan y,dx=\sec^2ydy$$
$$\int\sqrt{1+\frac1{\tan^2y}}\sec^2ydy=\int\sqrt{1+\cot^2y}\sec^2dy=\int\csc y\sec^2ydy$$
It will take a little work from here to get this to look like something recognizable.
$$\int\csc y(1+\tan^2y)dy=\int\csc ydy+\int\csc y\tan^2ydy$$
I assume you know how to do the first integral.  It's similar to the integral for $\sec y$.  As for the second half
$$\int\csc y\tan^2ydy=\int\frac1{\sin y}\times\frac{\sin y}{\cos y}\times\tan ydy=\int\sec y\tan ydy$$
which should now be in a form that looks familiar.
