Anti-differentiation

**sanado** · July 3rd 2008, 07:23 AM

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

**TKHunny** · July 3rd 2008, 07:33 AM

Try u = e^x - 1

Try the "Calculus" Section

**Moo** · July 3rd 2008, 07:37 AM

Hello,

Originally Posted by sanado

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 $x^4=(x-1)(x^3+x^2+x+1)+1$

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

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

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

For the last one, try $u=e^x-1$ and don't forget the constant of integration.

**flyingsquirrel** · July 3rd 2008, 07:55 AM

Hi

Originally Posted by Moo

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

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

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

@ sanado :

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

Letting $n=3$ and $u=\exp x$ it gives $1+\exp x +\exp (2x)+\exp(3x)=\frac{1-\exp (4x)}{1-\exp(x)}$ and from this you can easily find Moo's result : $\frac{\exp(4x)}{\exp x-1}=1+\exp x +\exp (2x)+\exp(3x)+\frac{1}{\exp x-1}$

**sanado** · July 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

**flyingsquirrel** · July 3rd 2008, 08:37 AM

Originally Posted by sanado

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 $\ln(\exp x-1)+\frac{\exp(3x)}{3}+\frac{\exp(2x)}{2}+\exp x+C$ which is the same as the answer. (hopefuly

) 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.

**TKHunny** · July 3rd 2008, 09:55 AM

Wow! That's a lot of work.

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

$u = e^{x}-1$

$du = e^{x}dx$

$u+1 = e^{x}$

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

It's a simple polynomial.

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