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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.

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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

$$ \begin{align} \int\frac{x}{(1+x^2)^{1/2}}dx&=\int\frac{\sinh u}{\cosh u}\:\cosh u \:du\\\\ &=\int\sinh u\:du\\\\ &=\cosh u+C\\\\ &=\sqrt{1+x^2}+C. \end{align} $$

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    $\begingroup$ Ah thank you, I see the problem is that when I took integral calculus, we didn't cover hyperbolic trig functions, but the textbook I'm using for ODEs seems to expect I either have some knowledge of them, or simply have an intuition of when it's the results of a trivial chain rule. $\endgroup$ – awiebe Oct 28 '15 at 1:27
  • $\begingroup$ @awiebe $\cosh u$ is just a shortcut for $(e^u+e^{-u})/2$, and many properties are not difficult to understand :) Thanks! $\endgroup$ – Olivier Oloa Oct 28 '15 at 1:29
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Hint: let $u=1+x^2$ then $du=\dots$.

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  • $\begingroup$ if u= 1+x^2 then du/dx =2x => du=2dx $$\int{x/u}du =do I have ro solve for x in terms of u to get rid of the other x? $\endgroup$ – awiebe Oct 28 '15 at 1:22
  • $\begingroup$ A typo: du=2xdx. So the "other x" is accounted for... $\endgroup$ – Barry Smith Oct 28 '15 at 2:38
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let, $x=\tan \theta\implies dx=\sec^2\theta \ d\theta$ $$\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$$$$=\sqrt{1+\tan^2\theta}+C$$ $$=\color{blue}{\sqrt{1+x^2}+C}$$

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