# Math Help - Partial fraction with substitution

1. ## 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?

2. 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

3. 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.

4. I got A = 2 and B = -1

5. 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.

6. 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

8. $\frac{u}{(u+2)(u+1)}=\frac{2(u+1)-(u+2)}{(u+2)(u+1)}=\frac{2}{u+2}-\frac{1}{u+1}.$
$\frac{u}{(u+2)(u+1)}=\frac{2(u+1)-(u+2)}{(u+2)(u+1)}=\frac{2}{u+2}-\frac{1}{u+1}.$