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Homework Help: Partial Fraction Integration

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I was told to solve using partial fractions. When I set up the partial fraction I got:

    A/e^x+1 + B/e^x-1 = 2e^x

    When I broke it up, I solved for A and B and got that A and B should both equal 1/2. When I plugged it back into the integral, I pulled out the 2 in the numerator, substituted A and B in (so the 2 and the 1/2s all cancel) and integrated 1/e^2x+1 dx and 1/e^2x-1 dx

    After I integrated, I got ln(e^x+1) + ln(e^x-1) + C

    However, when I looked up the answer online to check my answer, they had something a little different. The answer (on Wolfram Alpha) was nearly the same, but instead was: ln(1-e^x)-ln(e^x+1) +C

    I was wondering where I could have gone wrong?

    Just to be clear, when I set up the partial fractions, I did the following:

    A/e^x+1 + B/e^x-1 =2e^x/(e^x+1)(e^x-1).
    I set this equal to the numerator in the integral:
    (A+B)e^x+(B-A)=e^2x ---- I pulled the two outside of the integral in the first step
    This means:

    Any idea where I went wrong?
  2. jcsd
  3. Sep 10, 2013 #2


    Staff: Mentor

    1. Use parentheses. What you wrote on the left side is this:
    Ae-x + 1 + Be-x + 1

    What you meant, I'm sure, was this:
    A/(ex + 1) + B/(ex + 1)
    2. Your equation really would be this:
    A/(ex + 1) + B/(ex + 1) = 2ex/(e2x - 1)
    I suspect that where you went wrong was thinking this:

    $$\int \frac{dx}{e^x + 1} = ln(e^x + 1) + C$$
    That is NOT correct!!
    Furthermore, I think your approach is a dead end. A simpler approach is to do a substitution first, and then use partial fractions on that.
  4. Sep 10, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You wrote several incorrect formulas. Never, never write A/e^x+1 (which equals 1 + (A/e^x)) if you mean A/(e^x+1). Always use parentheses, or else use TeX, like this:
    [tex] \frac{A}{e^x+1}[/tex]

    Anyway, your answer makes sense only if x > 0 (so that e^x > 1) and the on-line answer makes sense only if x < 0 (so that e^x < 1); those restrictions are needed to avoid computing logs of negative numbers. You need to write things differently if you want a formula that works for all x, whether positive or negative.
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