# Integrate $\frac{x}{\sqrt{1+x^2}}$

I apologize if this is trivial, but I haven't had to do integral calculus in a while and I can't for the life of me remember how to find the indefinite integral

$$\int{\frac{x}{\sqrt{1+x^2}}}\,dx =?$$

I've tried a quick google and nothing came up. Wolfram tells me the

$$\int{\frac{x}{\sqrt{1+x^2}}} = \sqrt{x^2+1} +C$$

Which makes perfect sense to me if I differentiate it, but I cannot seem to remember how that can be determined backwards, apart from just memorizing.

You may use $$\cosh^2 u-\sinh^2 u=1$$ by setting $$x:=\sinh u$$ giving $$(1+x^2)^{1/2}=(1+\sinh^2 u)^{1/2}=(\cosh^2 u)^{1/2}=\cosh u.$$ Then
\begin{align} \int\frac{x}{(1+x^2)^{1/2}}dx&=\int\frac{\sinh u}{\cosh u}\:\cosh u \:du\\\\ &=\int\sinh u\:du\\\\ &=\cosh u+C\\\\ &=\sqrt{1+x^2}+C. \end{align}
• @awiebe $\cosh u$ is just a shortcut for $(e^u+e^{-u})/2$, and many properties are not difficult to understand :) Thanks! – Olivier Oloa Oct 28 '15 at 1:29
Hint: let $u=1+x^2$ then $du=\dots$.
• if u= 1+x^2 then du/dx =2x => du=2dx $$\int{x/u}du =do I have ro solve for x in terms of u to get rid of the other x? – awiebe Oct 28 '15 at 1:22 • A typo: du=2xdx. So the "other x" is accounted for... – Barry Smith Oct 28 '15 at 2:38 let, x=\tan \theta\implies dx=\sec^2\theta \ d\theta$$\int \frac{x}{\sqrt{1+x^2}}\ dx=\int \frac{\tan\theta}{(1+\tan^2\theta)^{1/2}}\ (\sec^2\theta \ d\theta)=\int \frac{\tan\theta\sec^2\theta }{(\sec^2\theta)^{1/2}}\ \ d\theta=\int \frac{\tan\theta\sec^2\theta}{\sec\theta}\ d\theta=\int \sec\theta\tan\theta\ d\theta=\int d(\sec\theta)=\sec\theta+C=\sqrt{1+\tan^2\theta}+C=\color{blue}{\sqrt{1+x^2}+C}