How to compute the integral of $ \frac{\sqrt{(x^2+1)}}{x^2}$? II have spent hours trying to find the right substitution but I have had no results.
I have tried using $x=\tan u$ but it did not give the result shown in my book.
 A: Using Integration by parts,
$$\int\dfrac{\sqrt{1+x^2}}{x^2}dx=\sqrt{1+x^2}\int\dfrac{dx}{x^2}-\int\left(\dfrac{d\sqrt{1+x^2}}{dx}\int\dfrac{dx}{x^2}\right)dx$$
$$=-\dfrac{\sqrt{1+x^2}}x+\int\dfrac{dx}{\sqrt{1+x^2}}$$
Hope you can take it from here!
A: As suggested in the comments, letting $x=\sinh u$ and exploiting $dx = \cosh u\, du$ as well as the identity $\cosh^2u-\sinh^2u=1$, we have
$$
\int \frac{\sqrt{1+x^2}}{x^2}dx=\int\frac{cosh^2 u}{\sinh^2u}du.
$$
Now,
$$
\coth u = \frac{\cosh u}{\sinh u}\implies \left( \coth u \right)'=\frac{\sinh^2 u - \cosh^2u}{\sinh^2 u} = 1-\frac{cosh^2 u}{\sinh^2u}.
$$
So
$$
\int \frac{\sqrt{1+x^2}}{x^2}dx = -\frac{\cosh u}{\sinh u} + u.
$$
In general, as a rule of thumb, you can think of hyperbolic functions as the natural substitution when encountering integrals of radicals of the following type
$$
\sqrt{x^2+a^2} \implies x=a\sinh u
$$
$$
\sqrt{x^2-a^2}\implies x=a\cosh u,
$$
in the same way as trigonometric functions help for radicals like
$$
\sqrt{a^2-u^2}\implies u=a\sin \theta.
$$
A: HINT:
Setting $x=\tan u$ $$\int\dfrac{\sqrt{x^2+1}}{x^2}dx =\int\dfrac{\sec u}{\tan^2u}\sec^2u\ du=\int\dfrac{du}{\cos u\sin^ 2u} =\int\dfrac{\cos u\ du}{\cos^2u\sin^ 2u}$$
Set $\sin u= v$
Hope you can take it from here!
A: $$\int \frac{\sqrt{x^2+1}}{x^2}dx$$
Apply Integration By Parts:$\:\int \:uv'=uv-\int \:u'v$
$u=\sqrt{x^2+1},\:u'=\frac{x}{\sqrt{x^2+1}},\:\:v'=\frac{1}{x^2},\:\:v=-\frac{1}{x}$
Then
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=-\frac{\sqrt{x^2+1}}{x}-\int \:-\frac{1}{\sqrt{x^2+1}}dx$$
$$=-\frac{\sqrt{x^2+1}}{x}-\int \:-\frac{1}{\sqrt{x^2+1}}dx$$
Then Note:
$\int \:-\frac{1}{\sqrt{x^2+1}}dx=-arcsinh \left(x\right)$
So
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=\color{red}{arcsinh \left(x\right)-\frac{\sqrt{x^2+1}}{x}+C}$$
