# Anti-differentiation

• Jul 3rd 2008, 07:23 AM
Anti-differentiation
Hey guys, sorry if this question is posted in the wrong section but i didnt no where else to submit it. Anywayz, the question is:

Find the anti-derivative of

e^4x / e^x - 1

Ive looked at the question, tried a partial fraction, tried the change of variable method but cant get close to the answer. Working out would be much appreciated. Thankz
• Jul 3rd 2008, 07:33 AM
TKHunny
Try u = e^x - 1

Try the "Calculus" Section
• Jul 3rd 2008, 07:37 AM
Moo
Hello,

Quote:

Hey guys, sorry if this question is posted in the wrong section but i didnt no where else to submit it. Anywayz, the question is:

Find the anti-derivative of

e^4x / e^x - 1

Ive looked at the question, tried a partial fraction, tried the change of variable method but cant get close to the answer. Working out would be much appreciated. Thankz

Hmm well, partial fraction would have worked...

Note that $\displaystyle x^4=(x-1)(x^3+x^2+x+1)+1$

Therefore $\displaystyle \frac{e^{4x}}{e^x-1}=\frac{(e^x-1)(e^{3x}+e^{2x}+e^x+1)+1}{e^x-1}$

$\displaystyle =e^{3x}+e^{2x}+e^x+1+\frac{1}{e^x-1}$

--> $\displaystyle \int \frac{e^{4x}}{e^x-1} \, dx=\int e^{3x} \, dx+\int e^{2x} \, dx+\int e^x \, dx+\int \, dx+\int \frac{1}{e^x-1} \, dx$

For the last one, try $\displaystyle u=e^x-1$ and don't forget the constant of integration.
• Jul 3rd 2008, 07:55 AM
flyingsquirrel
Hi
Quote:

Originally Posted by Moo
Note that $\displaystyle x^4=(x-1)(x^3+x^2+x+1)+1$

Therefore $\displaystyle \frac{e^{4x}}{e^x-1}=\frac{(e^x-1)(e^{3x}+e^{2x}+e^x+1)+1}{e^x-1}$

$\displaystyle =e^{3x}+e^{2x}+e^x+1+\frac{1}{e^x-1}$

To find this you can think about the sum of the terms of a geometric sequence : $\displaystyle 1+u+u^2+\ldots+u^n=\frac{1-u^{n+1}}{1-u}$

Letting $\displaystyle n=3$ and $\displaystyle u=\exp x$ it gives $\displaystyle 1+\exp x +\exp (2x)+\exp(3x)=\frac{1-\exp (4x)}{1-\exp(x)}$ and from this you can easily find Moo's result : $\displaystyle \frac{\exp(4x)}{\exp x-1}=1+\exp x +\exp (2x)+\exp(3x)+\frac{1}{\exp x-1}$
• Jul 3rd 2008, 07:57 AM
Hmmmmm ive tried the method above but i aint getting the answer which is:

1/3.(e^x - 1)^3 + 3/2.(e^x - 1)^2 + 3(e^x - 1) + loge(e^x - 1) + c
• Jul 3rd 2008, 08:37 AM
flyingsquirrel
Quote:

Hmmmmm ive tried the method above but i aint getting the answer which is:

1/3.(e^x - 1)^3 + 3/2.(e^x - 1)^2 + 3(e^x - 1) + loge(e^x - 1) + c

The method above should give you $\displaystyle \ln(\exp x-1)+\frac{\exp(3x)}{3}+\frac{\exp(2x)}{2}+\exp x+C$ which is the same as the answer. (hopefuly :D) If it still does not correspond to what you've found, show us what you've done so that we can try to find what the problem is.
• Jul 3rd 2008, 09:55 AM
TKHunny
Wow! That's a lot of work.

$\displaystyle \int \frac{e^{4x}}{e^{x}-1}\;dx$

$\displaystyle u = e^{x}-1$

$\displaystyle du = e^{x}dx$

$\displaystyle u+1 = e^{x}$

$\displaystyle \int \frac{(u+1)^{3}}{u}\;du$

It's a simple polynomial.