∫dx/(1-e^x) I have no idea what to set u equal to. Could someone help me begin this problem?
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Originally Posted by r2d2 $\displaystyle ∫dx/(1-e^x)$ $\displaystyle \int\frac{dx}{1-e^x}=\int\frac{dx}{1-e^x}\cdot\frac{e^{-x}}{e^{-x}}=\int\frac{e^{-x}}{e^{-x}-1}\text{ }dx$...so
Ok, So I'm stuck on why you multpilied by $\displaystyle e^(-x)/e^(-x)$
Originally Posted by r2d2 Ok, So I'm stuck on why you multpilied by $\displaystyle e^(-x)/e^(-x)$ Let $\displaystyle u=....$. That's why.
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)?
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
great, so i just need to add the negative, and the c, then re-substitute the u function. Thanks, mate.
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