# integration by substitution

• Jul 20th 2011, 06:58 AM
Punch
integration by substitution
$\displaystyle \int\frac{1}{e^x+1}dx ($let $\displaystyle u=e^x+1)$

$\displaystyle \int\frac{1}{e^x+1}dx=\int\frac{1}{u}\frac{1}{u-1}du$

$\displaystyle =\int\frac{1}{u^2-u}du$

$\displaystyle =\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du$

$\displaystyle =ln(\frac{u-1}{u})+c$

$\displaystyle =ln(\frac{e^x}{e^x+1})+c$
• Jul 20th 2011, 07:02 AM
Siron
Re: integration by substitution
You can wright the numerator as:
$\displaystyle \int \frac{1}{e^x+1}dx= \int \frac{(e^x+1)-(e^x)}{e^x+1}dx=\int \frac{e^x+1}{e^x+1}dx - \int \frac{e^xdx}{e^x+1}$
Now use the substitution $\displaystyle t=e^x+1$ for the second integral.
• Jul 20th 2011, 07:02 AM
Prove It
Re: integration by substitution
Quote:

Originally Posted by Punch
$\displaystyle \int\frac{1}{e^x+1}dx ($let $\displaystyle u=e^x+1)$

$\displaystyle \int\frac{1}{e^x+1}dx=\int\frac{1}{u}\frac{1}{u-1}du$

$\displaystyle =\int\frac{1}{u^2-u}du$

$\displaystyle =\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du$

$\displaystyle =ln(\frac{u-1}{u})+c$

$\displaystyle =ln(\frac{e^x}{e^x+1})+c$

This is the same mistake you made in the last question. If there is a u^2 term in the denominator, the integral is NOT a logarithm unless there is a u term in the numerator.

I would start by multiplying top and bottom by e^x...
• Jul 20th 2011, 11:02 AM
Jester
Re: integration by substitution
Here's another way

$\displaystyle \int \dfrac{1}{e^x+1}dx = \int \dfrac{e^{-x}}{1 + e^{-x}}dx$

and let $\displaystyle u = 1 + e^{-x}$