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I'm working with

$$ \int \frac{e^x}{(e^x - 1)(e^x + 2)}dx $$

So, I know I'll be doing a u-sub and a partial fraction decomposition.

I'll let
$$ u = e^x $$

Making my equation $$ \int \frac{u}{(u - 1)(u + 2)} $$

Then I do my partial fraction decomposition.

$$ u = A(u + 2) + B(u - 1) $$

Let u = -2, $$ B(-2 - 1) = -2$$ therefore $$B = \frac{2}{3} $$

Let u = 1, $$ A(1 + 2) = 1$$ therefore $$ A = \frac{1}{3} $$

This leaves me with

$$ \int \frac{u}{3(u - 1)} + \frac{2u}{3(u + 2)} $$

Where do I go from here? Do I plug $e^x$ back in for u and cancel out like terms? If I do that, won't I have to worry about the bottom of the fraction possibly being 0?

Edit: Reworking without dropping the du.

Let $u = e^x$ and $du = e^x dx$

Giving me $$ \int \frac{du}{(u-1)(u+2)} $$

So, if I split these, I have

$$ \int \frac{du}{3(u-1)} + \int \frac{2du}{3(u+2)} $$

This gives me

$$ \int \frac{1}{3(u-1)} + \int \frac{2}{3(u+2)} $$

So for the first part, I'll let $w = u - 1$ and bring the $\frac{1}{3}$ out front

$$ \frac{1}{3} \int \frac{dw}{w} $$

This gives me

$$ \frac{1}{3} ln | w | $$

which ultimately gives me

$$ \frac{1}{3} ln | e^x - 1 | $$

Do the same for the other side, let $w = u + 2$ and bring the fraction out front

$$ \frac{2}{3} \int \frac{dw}{w} $$

which gives me

$$ \frac{2}{3} ln | e^x + 2 | $$

put them together and my final answer is

$$ \frac{1}{3} ln | e^x - 1 | - \frac{1}{3} ln | e^x + 2 | + c $$

Does that seem correct? - Revised & correct now :)

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  • 1
    $\begingroup$ Check your substitution again. $\endgroup$ – Potato Sep 25 '13 at 3:10
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    $\begingroup$ Just to be picky, what happened to $du$? $\endgroup$ – DJohnM Sep 25 '13 at 3:11
  • $\begingroup$ And what happened to the "2" in the value of B?? $\endgroup$ – DJohnM Sep 25 '13 at 3:16
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When you take $u=e^x$, then $du=e^x\ dx$.

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  • $\begingroup$ I ended up falling asleep last night before I was able to finish it. I put my edited work up on the main post if you'd like to check it out, I would appreciate it :) $\endgroup$ – ConfusingCalc Sep 25 '13 at 17:41
  • $\begingroup$ Indeed, since it is an indefinite integral :) thanks Danny! Hmm wolfram doesn't have a $\frac{2}{3}$ but rather a $\frac{1}{3}$ I can't quite seem to see how my B is suppose to be $\frac{1}{3}$ though $\endgroup$ – ConfusingCalc Sep 25 '13 at 17:58
  • $\begingroup$ I'm not quite sure I understand how it should be -1 instead of -2 though. I thought I had to plug in -2 into u to zero A out? $\endgroup$ – ConfusingCalc Sep 25 '13 at 18:15
  • $\begingroup$ So wolfram is wrong? Hehe I trust you & wolfram more than myself, it says the answer is: $$ \frac{1}{3}(log(1-e^x) - log(e^x +2)) + c $$ $\endgroup$ – ConfusingCalc Sep 25 '13 at 18:42
  • $\begingroup$ Note that you let $u=e^x$ so $du=e^x dx$, therefore there shouldn't be an $u$ on the top! You should get $\frac1{(u-1)(u+2)}$ instead of $\frac u{(u-1)(u+2)}$ $\endgroup$ – user67258 Sep 25 '13 at 18:45
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HINT 1: When you set $u=e^x$ then $du=e^xdx$. Meaning that what you have to integrate is:

$$ \int \frac{du}{(u-1)(u+2)} $$

HINT 2: Use partial fractions to show that:

$$ \frac{1}{(u-1)(u+2)} = \frac{1}{3(u-1)} - \frac{1}{3(u+2)} $$

so you can independetly integrate each term and then subtract.

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  • $\begingroup$ So instead of setting my partial fractions equal to 1, they should be equal to du? $\endgroup$ – ConfusingCalc Sep 25 '13 at 3:37
  • $\begingroup$ Instead of setting your partial fractions equal to $u$ it should be equal to $1$. The $du$ is in some sense like a $1$. It means only your integrating with respect to $u$. Every integral must have a $d$ something. Here, every integral after separating in partial fractions will have its own $du$. $\endgroup$ – Mauricio G Tec Sep 25 '13 at 4:44
  • $\begingroup$ So, if I evaluate the first integral, $\int \frac{1}{3(u-1)}$ I'll let $w = u-1$ so I have $\frac{1}{3} \int \frac{dw}{w}$ giving me $\frac{1}{3} ln | w |$ which is $\frac{1}{3} ln | u-1 |$ ultimately giving me $\frac{1}{3} ln |e^x -1 |$ for the first one. Assuming this is correct, I do the second one the same way. I'll post my revised answer in the main post, if you could check it, that'd be great. $\endgroup$ – ConfusingCalc Sep 25 '13 at 17:33
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I think we have to go back to the basics.

Remember int(x+2)DX the DX multiplies through the individual terms!

So when u decomposed a product into sums. The DX passes through each term! A subtle observation which is assumed by calc teachers!

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