# perform the integration

• Apr 19th 2009, 03:06 PM
venikUSA
perform the integration
Can someone show me the solution to this problem? I have the exam tomorrow :(

the integral of e^-3x/12+e^-3x

Thank you!!!!
• Apr 19th 2009, 03:22 PM
derfleurer
So integrate (e^-3x)/(12 + e^-3x)?

What does our natural log rule say? u'/u = ln|u| + c, right? Like the integral of 1/x is ln|x|. You've got the same thing here. Only difference is, u' would equal -3e^-3x. Well, we have -1/3 of that. Fix this by throwing a -1/3 outside the integral sign.

Then it's just $-1/3 \int u'/u$. Or -1/3ln|u| + c
• Apr 19th 2009, 03:22 PM
skeeter
Quote:

Originally Posted by venikUSA
Can someone show me the solution to this problem? I have the exam tomorrow :(

the integral of e^-3x/12+e^-3x

Thank you!!!!

e^(-3x)/[12+e^(-3x)]

in future, use grouping symbols to make your expressions clear ... or learn how to use latex

$\int \frac{e^{-3x}}{12 + e^{-3x}} \, dx$

$-\frac{1}{3} \int \frac{-3e^{-3x}}{12 + e^{-3x}} \, dx$

$-\frac{1}{3} \ln(12 + e^{-3x}) + C$

btw, you don't need the absolute value in this particular case ... why?