$\displaystyle \int\frac{e^{2x}}{e^{2x}+3e^x+2}dx$

I made the substitution: $\displaystyle u=e^x$ so $\displaystyle du=e^xdx$

This got me the wrong answer (though my math may have been wrong), is this the correct substitution to make?

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- Sep 9th 2008, 08:18 PMsymstarPartial fraction with substitution
$\displaystyle \int\frac{e^{2x}}{e^{2x}+3e^x+2}dx$

I made the substitution: $\displaystyle u=e^x$ so $\displaystyle du=e^xdx$

This got me the wrong answer (though my math may have been wrong), is this the correct substitution to make? - Sep 9th 2008, 08:27 PMChris L T521
- Sep 9th 2008, 10:06 PMsymstar
Yes and from there I got:

$\displaystyle \frac{u}{(u+2)(u+1)}=\frac{A}{u+2}+\frac{B}{u+1} \rightarrow u=A(u+1)+B(u+2)$

$\displaystyle u=-2 \text{ then } A=-2$

$\displaystyle u=-1 \text{ then } B=-1$

So...

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

$\displaystyle -2ln\left|u+2\right|-ln\left|u+1\right|+C$

...and that doesn't seem to be leading to the correct answer, so I think I went wrong somewhere in there. - Sep 9th 2008, 10:13 PM11rdc11
I got A = 2 and B = -1

- Sep 9th 2008, 10:13 PMMoo
- Sep 9th 2008, 10:35 PM11rdc11
Another way to find A and B is

u = Au + A + Bu +2B

Now gather the coefficents for each power and make and equation, for example

1 = A + B for power of 1

0 = A + 2B for power of 0

Now you can use substitution or elimination to get your A and B values - Sep 10th 2008, 05:57 AMsymstar
Thanks for your help!

- Sep 10th 2008, 09:13 PMKrizalid
$\displaystyle \frac{u}{(u+2)(u+1)}=\frac{2(u+1)-(u+2)}{(u+2)(u+1)}=\frac{2}{u+2}-\frac{1}{u+1}.$

- Sep 10th 2008, 09:21 PMChris L T521