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.


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



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 $u=\sqrt{x+1}$

  • $\begingroup$ nice..........................+1 $\endgroup$ – Bhaskara-III Jan 11 '17 at 6:09

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.

  • $\begingroup$ What is the mathjax for argth ? I had to \operatorname it in this answer. $\endgroup$ – zwim Jan 11 '17 at 6:14
  • $\begingroup$ what is argth? hyperbolic function i presume? $\endgroup$ – bigfocalchord Jan 11 '17 at 6:16
  • $\begingroup$ Reverse of hyperbolic tangent, the equivalent of arctan for hyperbolic stuff. $\endgroup$ – zwim Jan 11 '17 at 6:18

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