# Partial fraction with substitution

• Sep 9th 2008, 08:18 PM
symstar
Partial fraction with substitution
$\int\frac{e^{2x}}{e^{2x}+3e^x+2}dx$

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

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 PM
Chris L T521
Quote:

Originally Posted by symstar
$\int\frac{e^{2x}}{e^{2x}+3e^x+2}dx$

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

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

That is the correct substitution. Did you get this for your integral?

$\int\frac{u\,du}{u^2+3u+2}\implies\int\frac{u\,du} {(u+2)(u+1)}$

--Chris
• Sep 9th 2008, 10:06 PM
symstar
Yes and from there I got:

$\frac{u}{(u+2)(u+1)}=\frac{A}{u+2}+\frac{B}{u+1} \rightarrow u=A(u+1)+B(u+2)$
$u=-2 \text{ then } A=-2$
$u=-1 \text{ then } B=-1$

So...

$\int\frac{-2}{u+2}-\frac{1}{u+1}du$
$-2ln\left|u+2\right|-ln\left|u+1\right|+C$

...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 PM
11rdc11
I got A = 2 and B = -1
• Sep 9th 2008, 10:13 PM
Moo
Quote:

Originally Posted by symstar
Yes and from there I got:

$\frac{u}{(u+2)(u+1)}=\frac{A}{u+2}+\frac{B}{u+1} \rightarrow u=A(u+1)+B(u+2)$
$u=-2 \text{ then } A=-2$
$u=-1 \text{ then } B=-1$

So...

$\int\frac{-2}{u+2}-\frac{1}{u+1}du$
$-2ln\left|u+2\right|-ln\left|u+1\right|+C$

...and that doesn't seem to be leading to the correct answer, so I think I went wrong somewhere in there.

Here is the problem.

If u=-2, then -2=A*(-2+1)=-A
Thus A=2.
• Sep 9th 2008, 10:35 PM
11rdc11
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 AM
symstar

• Sep 10th 2008, 09:13 PM
Krizalid
$\frac{u}{(u+2)(u+1)}=\frac{2(u+1)-(u+2)}{(u+2)(u+1)}=\frac{2}{u+2}-\frac{1}{u+1}.$
• Sep 10th 2008, 09:21 PM
Chris L T521
Quote:

Originally Posted by Krizalid
$\frac{u}{(u+2)(u+1)}=\frac{2(u+1)-(u+2)}{(u+2)(u+1)}=\frac{2}{u+2}-\frac{1}{u+1}.$

Aha! That's the trick I couldn't remember! :D

--Chris