# Thread: Integration with u sub

1. ## Integration with u sub

integrate (-4-4e^x)/(1-e^x) using u substitution where u=e^x

I've done it a few ways and gotten...
4ln(1-e^-x)+4ln(1-e^x)+c
8ln(1-e^x)-4x+c
4ln(e^x)+4ln(1-e^x)+4ln(e^x-1)+c

neither of which are correct, any insight on what I'm doing wrong?

2. ## Re: Integration with u sub

Hello, Johngalt13!

$\text{Integrate: }\:\int\frac{-4-4e^x}{1-e^x}\,dx\:\text{ using u-substitution: }u\,=\,e^x$

I've done it a few ways and gotten...
. . $4\ln|1 - e^{-x}| + 4\ln|1 - e^x| + C$
. . $8\ln|1 - e^x| - 4x + C$ **
. . $4\ln(e^x) + 4\ln|1 - e^x| + 4\ln|e^x - 1| + C$

I solved it like this . . .

$\int\frac{-4-4e^x}{1-e^x}\,dx \;=\;-4\int\frac{1+e^x}{1-e^x}\,dx \;=\;4\int\frac{e^x+1}{e^x-1}\,dx$

$\text{Let }u = e^x \quad\Rightarrow\quad x = \ln u \quad\Rightarrow\quad dx = \frac{du}{u}$

$\text{Substitute: }\:4\int \frac{u+1}{u-1}\,\frac{du}{u} \;=\;4\int\frac{u+1}{u(u-1)}\,dx$

$\text{Partial Fractions: }\:4\int\frac{u+1}{u(u-1)}\,du \;=\;4\int\left(\frac{2}{u-1} - \frac{1}{u}\right)du$

. . . . . . . . . . . $=\;4\big(2\ln|u-1| - \ln|u|\big) + C$

$\text{Back-substitute: }\:4\big(2\ln|e^x-1| - \ln|e^x|\big) + C \;=\;$

. . . . . . . . . . $=\;4\big(2\ln|e^x-1| - x\big) + C$

. . . . . . . . . . $=\;8\ln|e^x-1| - 4x + C$

$\text{Note that: }\:|e^x - 1| \:=\:|1-e^x|$