# what did i do wrong? Integral problem:

• Dec 12th 2009, 06:07 PM
micro31337
what did i do wrong? Integral problem:
Am i doing this problem wrong?
Evaluate the integral
$\displaystyle \int (e^(2x)) / (e^x - 2)) dx$ <-- that is e^(2x)
i expanded the problem to:
$\displaystyle \int(e^x * e^x * (1/ (e^x - 2)) dx$

i picked u to be e^x - 2
$\displaystyle u = e^x - 2, e^x = u + 2$
$\displaystyle du = e^x$
rewrite to:
$\displaystyle \int((u + 2) * 1/ (u) dx$

multiply all out to
$\displaystyle \int 1 + 2*(1/u) dx$

integrate and get:
$\displaystyle u + 2ln|u| + C$
$\displaystyle (e^x - 2) + 2ln|e^x - 2|+C$

$\displaystyle e^x + 2ln|e^x - 2|+C$

what did i do wrong?
• Dec 12th 2009, 06:18 PM
oblixps
you didn't do anything wrong. your answer is correct. notice that the -2 from (e^x - 2) can be combined with the constant C to produce another arbitrary constant. once you do that, your answer will look just like the other answer.
• Dec 12th 2009, 06:22 PM
micro31337
you can combine it with the constant? Hmm never done that before, thanks.
• Dec 12th 2009, 06:34 PM
Soroban
Hello, micro31337!

You did nothing wrong . . .

Quote:

Integrate and get: .$\displaystyle (e^x - 2) + 2\ln|e^x - 2|+C$

but the answer says it's: .$\displaystyle e^x + 2\ln|e^x - 2|+C$

$\displaystyle \text{You have: }\;e^x - 2 \;+\; 2\ln|e^x-2| \;+\; C \;\;=\;\;e^x \;+\; 2\ln|e^x-2| \;+\; \underbrace{C - 2}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
This is an arbitrary constant

Therefore: . $\displaystyle e^x + 2\ln|e^x-2| + C$