# Partial Fraction Integration

1. Sep 10, 2013

### Yosty22

1. The problem statement, all variables and given/known data

∫(2e^x)/(e^(2x)-1)dx

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).
=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?

2. Sep 10, 2013

### Staff: Mentor

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

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.

3. Sep 10, 2013

### Ray Vickson

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:
$$\frac{A}{e^x+1}$$

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.