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I am trying to integrate this function:

$$\int {\frac{1}{x\sqrt{x+1}}\mathrm{d}x}.$$

When I googled it I saw methods that used both "$u$" and "$s$" substitution. I sort of understood what was going on but got stuck after substituting in $s$. I could not simplify any further.

My professor used a really weird substitution where she found $u^2$ and then declared: $\frac{2u(du)}{(u+1)(u-1)}$ then used the heavyside method.I understood her use of the heavyside method however I do not understand where her above function came from. What happened to $x$? How did $2u$ get on top?

Could you please help me understand how both of these methods could be used to solve this integral? It would be especially helpful to know where $\frac{2u(du)}{(u+1)(u-1)}$ came from.

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4 Answers 4

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Take $u^{2}=x+1$ so that $2udu=dx$ and $u^{2}-1=x$. This was done to remove the square root. Then we get

$$\int\frac{2u}{u(u^{2}-1)}du=\int\frac{2}{(u-1)(u+1)}du=\int\left(\frac{1}{u-1}-\frac{1}{u+1}\right)du$$

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Put $x+1=u^2$ and $dx=2udu$ to get $$\int \frac{2du}{u^2-1}$$$=log\vert (u-1)\vert-log\vert (u+1)\vert+c$ where $u=\sqrt{x+1}$

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  • $\begingroup$ nice..........................+1 $\endgroup$ Jan 11, 2017 at 6:09
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Remark there is not real need to separate $(u^2-1)$ into $(u-1)(u+1)$ since $\int\frac{du}{1-u^2}=\operatorname{argth}(u)$.

Your integral is $-2\operatorname{argth}(\sqrt{x+1})$ which of course is equivalent to the log formulation.

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  • $\begingroup$ What is the mathjax for argth ? I had to \operatorname it in this answer. $\endgroup$
    – zwim
    Jan 11, 2017 at 6:14
  • $\begingroup$ what is argth? hyperbolic function i presume? $\endgroup$ Jan 11, 2017 at 6:16
  • $\begingroup$ Reverse of hyperbolic tangent, the equivalent of arctan for hyperbolic stuff. $\endgroup$
    – zwim
    Jan 11, 2017 at 6:18
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$$ \begin{aligned} \int \frac{1}{x \sqrt{x+1}} & =2 \int \frac{d(\sqrt{x+1})}{(\sqrt{x+1})^2-1} \\ & =\ln \left|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\right|+C \end{aligned} $$

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