Hi, got a question about the intergral of e^x+1/e^x-1
I put u=e^x-1, dx=du/e^x
And i am left with the intergral of 1/u+1, which isnt right. Can anyone help me?
Yes, if you let $\displaystyle u= e^x- 1$ then $\displaystyle du= e^xdx$ so that $\displaystyle dx= du/e^x= du/(u+ 1)$.
Also $\displaystyle e^x+ 1= u+ 2$.
But then $\displaystyle \int \frac{e^x+ 1}{e^x- 1}dx= \int \frac{u+ 2}{u}(u+1)du= \int\frac{(u+2)(u+1)}{u}du$ which is certainly NOT "1/(u+1)".
Yes, I think that was a typo/slip. Just in case a picture helps - in confirming your reasoning re the typo but then also getting past your alleged 1/(u+1)...
... where (key in spoiler) ...
Spoiler:
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!