Question

find the integral (e^x) / ((e^x)-1) dx

Answer 1

\[\large \int\limits \frac{e^x\;dx}{e^x-1}\]

Apply a `U-substitution`. Let your `u` be the denominator.

Answer 2

if u=e^x -1 the du/dx= e^x

Answer 3

Good, now move the dx to the other side c: giving you, du= e^x dx

Answer 4

Understand how you'll plug those pieces in? :D

Answer 5

kind of... would it be...the integral of e^x du/dx?

Answer 6

no you get e^x du = e^x e^x dx

Answer 7

is that right? or am i getting it confused?

Answer 8

Ah sorry, I got distracted.

Answer 9

\[\large \color{royalblue}{u=e^x-1} \qquad \qquad \color{salmon}{du=e^x \; dx}\]

Just switch the blue with blue, and the pink with pink to get them plugged in correctly.
\[\large \int\limits\limits \frac{\color{salmon}{e^x\;dx}}{\color{royalblue}{e^x-1}}\]

Answer 10

\[\large \int\limits\limits \frac{\color{salmon}{e^x\;dx}}{\color{royalblue}{e^x-1}} \qquad \rightarrow \qquad \int\limits\limits \frac{\color{salmon}{du}}{\color{royalblue}{u}}\]

Answer 11

does that make sense? :)

Answer 12

up to that point yes....but then do im confused as to how to continue