How to compute the integral of $ \frac{\sqrt{(x^2+1)}}{x^2}$?

Question

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.

Answer 1

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!

Answer 2

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!

Answer 3

$$\int \frac{\sqrt{x^2+1}}{x^2}dx$$ 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 $$\int \frac{\sqrt{x^2+1}}{x^2}dx=\color{red}{arcsinh \left(x\right)-\frac{\sqrt{x^2+1}}{x}+C}$$

Answer 4

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 $$ \sqrt{a^2-u^2}\implies u=a\sin \theta. $$