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Math Help - Partial fraction with substitution

  1. #1
    Junior Member symstar's Avatar
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    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?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by symstar View Post
    \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
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  3. #3
    Junior Member symstar's Avatar
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    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.
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  4. #4
    Super Member 11rdc11's Avatar
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    I got A = 2 and B = -1
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  5. #5
    Moo
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    Quote Originally Posted by symstar View Post
    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.
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  6. #6
    Super Member 11rdc11's Avatar
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    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
    Last edited by 11rdc11; September 10th 2008 at 01:26 AM.
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  7. #7
    Junior Member symstar's Avatar
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    Thanks for your help!

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  8. #8
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    \frac{u}{(u+2)(u+1)}=\frac{2(u+1)-(u+2)}{(u+2)(u+1)}=\frac{2}{u+2}-\frac{1}{u+1}.
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  9. #9
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Krizalid View Post
    \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!

    --Chris
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