0
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $u=\sqrt{x+1}$

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

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.