- #1

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## Homework Statement

∫(2e^x)/(e^(2x)-1)dx

## Homework Equations

## 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).

=A(e^x-1)+B(e^x+1)

=Ae^x-A+Be^x+B

=(A+B)e^x+(B-A)

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:

A+B=1

B-A=0

A=B=0.5

Any idea where I went wrong?