Integral Problem!

**r2d2** · January 27th 2010, 04:48 PM

∫dx/(1-e^x)

I have no idea what to set u equal to.

Could someone help me begin this problem?

**Drexel28** · January 27th 2010, 04:49 PM

Originally Posted by r2d2

$∫dx/(1-e^x)$

$\int\frac{dx}{1-e^x}=\int\frac{dx}{1-e^x}\cdot\frac{e^{-x}}{e^{-x}}=\int\frac{e^{-x}}{e^{-x}-1}\text{ }dx$ ...so

**r2d2** · January 27th 2010, 04:54 PM

Ok, So I'm stuck on why you multpilied by $e^(-x)/e^(-x)$

**Drexel28** · January 27th 2010, 04:55 PM

Originally Posted by r2d2

Ok, So I'm stuck on why you multpilied by $e^(-x)/e^(-x)$

Let $u=....$ . That's why.

**r2d2** · January 27th 2010, 05:00 PM

so setting u=e^(-x), I get the integral of [du/(e^(-x)-1)]-1. Would that be correct?

Then the integral of 1/(u-1)du = ln(u-1)?

**Drexel28** · January 27th 2010, 05:01 PM

Originally Posted by r2d2

so setting u=e^(-x), I get the integral of [du/(e^(-x)-1)]-1. Would that be correct?

Then the integral of 1/(u-1)du = ln(u-1)?

You forgot a negative...but

**r2d2** · January 27th 2010, 05:02 PM

great, so i just need to add the negative, and the c, then re-substitute the u function.

Thanks, mate.

Thread: Integral Problem!