# Math Help - integration by parts

1. ## integration by parts

HI

calculating integral sign ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

So if u = x then du = dx

then I need a dv .... if dv = e^-1 then v = e^-1 doesn't it? and how does this help me if I now introduce V?

how can I get all of these "e"s into "u"s?

Thanks

Calculus Beginner

2. Originally Posted by calcbeg
HI

calculating integral sign ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

So if u = x then du = dx

then I need a dv .... if dv = e^-1 then v = e^-1 doesn't it? and how does this help me if I now introduce V?

how can I get all of these "e"s into "u"s?

Thanks

Calculus Beginner
did you mean ...

$\int_0^2 \frac{e^{-x}}{1-e^{-2x}} \, dx$

or what you posted?

3. what I posted - there is an x then the e stuff

4. Originally Posted by calcbeg
what I posted - there is an x then the e stuff
your notation is a bit confusing ...

("S") b=2; a=0 (e^-x/(1-e^-2)) dx

is the integrand

$\frac{e^{-x}}{1-e^{-2}}$

or is it

$e^{\frac{-x}{1-e^{-2}}}$

5. If it really is, as you said, ("S") b=2; a=0 (e^-x/(1-e^-2)) dx
which is $\int\frac{e^{-x}}{1- e^{-2}}dx$, then that is the same as $\frac{1}{1- e^{-1}}\int e^{-x}dx$ and that should be easy to integrate.

6. Originally Posted by HallsofIvy
If it really is, as you said, ("S") b=2; a=0 (e^-x/(1-e^-2)) dx
which is $\int\frac{e^{-x}}{1- e^{-2}}dx$, then that is the same as $\frac{1}{1- e^{-1}}\int e^{-x}dx$ and that should be easy to integrate.
Why did you take out $\frac{1}{1- e^{-1}}$? Wouldn't the e be to the negative 2 power-- $\frac{1}{1- e^{-2}}$ since all of it is a constant?