Question

Guys, is there someone out there kind enough to help me with my calculus assignment? The idea is to share knowledge and techniques about integration methods, for instance, how would you integrate dx/sqrt((e^x)-1)?

Answer 1

Hmmm \[
\int \frac{dx}{\sqrt{e^x-1}}dx
\]

Answer 2

Do you know how to do u substitutions, integration by parts, and trig functions?

Answer 3

yeah, I do, but i wanna know, how would you solve it?

Answer 4

\[\int\frac{1}{\sqrt{e^x-1}}dx\]
\[u=e^x\\
du=e^xdx\iff \frac{1}{u}du=dx\]
\[\int\frac{1}{u\sqrt{u-1}}du\]
\[t=u-1\iff u=1+t\\
dt=du\]
\[\int\frac{1}{(1+t)\sqrt{t}}dt\]
\[v=\sqrt{t}\\
dv=\frac{1}{2\sqrt{t}}dt\iff 2v\;dv=dt\]
\[2\int\frac{v}{(1+v^2)v}dv\\
2\int\frac{1}{1+v^2}dv\]

Answer 5

D= I don't know why I hadn't seen this, thank you very much! After these days, i also came up with a solution:

\[u^2=e^x-1\]
\[e^x=u^2+1 => x=\ln(u^2+1)\]
\[dx=\frac{ 2udu }{ u^2+1 }\]
Now, doing the substitution:
\[\int\limits \frac{ \frac{ 2udu }{ u^2+1 } }{ u }du=2\int\limits \frac{ du }{ u^2+1 }\]

Which leads to the same solution! What do you think? Thank you very much for your help, and sorry I hadn't noticed until now @SithsAndGiggles

Answer 6

I made a mistake, I accidentally wrote two du when doing the substitution, don't pay attention to the second one

Answer 7

Hey, if it works, it works. I prefer rewriting (multiple times if I have to) until I get nicer-looking integral. You're welcome.