# Thread: what did i do wrong? Integral problem:

1. ## what did i do wrong? Integral problem:

Am i doing this problem wrong?
Evaluate the integral
$
\int (e^(2x)) / (e^x - 2)) dx$
<-- that is e^(2x)
i expanded the problem to:
$\int(e^x * e^x * (1/ (e^x - 2)) dx$

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

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

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

but the answer says its:
$e^x + 2ln|e^x - 2|+C$

what did i do wrong?

2. 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.

3. you can combine it with the constant? Hmm never done that before, thanks.

4. Hello, micro31337!

You did nothing wrong . . .

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

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

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

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