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Integration problem

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{2e^{2t}*dt}{e^{4t}+12e^{2t}+32}[/tex]

    3. The attempt at a solution
    I factored the denominator, getting (e2t+4)(e2t+8), from that I was able to express the original equation in terms of two fractions:
    [tex]\int \frac{A}{e^{2t}+4}[/tex]+[tex]\frac{B}{e^{2t}+8}[/tex]

    I then found A to be -2 and B=4.

    But when I put those back into the original equations, I still have integrals that have an 'e' on the bottom and I don't know what to do next...

    Any help is appreciated.
  2. jcsd
  3. Jan 23, 2009 #2


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    Put u=e2t, and then split into partial fractions.
  4. Jan 23, 2009 #3
    More partial fractions? But I can't. u+4 doesn't factor...?
  5. Jan 23, 2009 #4


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    \int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt

    u=e^2t => du/2 =e^2t dt

    [tex]\int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt \equiv \frac{1}{2} \int \frac{1}{u^2+12u+32} du[/tex]

    Now do partial fractions.
  6. Jan 23, 2009 #5
    I already did. It's in my procedure...
  7. Jan 23, 2009 #6


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    Then you'd simply have to integrate

    [tex]\frac{1}{8} \int (\frac{1}{u+4} - \frac{1}{u+8})du[/tex]
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