# integration by parts

• Oct 18th 2009, 12:51 PM
calcbeg
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
• Oct 18th 2009, 01:00 PM
skeeter
Quote:

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?
• Oct 18th 2009, 01:11 PM
calcbeg
what I posted - there is an x then the e stuff
• Oct 18th 2009, 01:16 PM
skeeter
Quote:

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

$e^{\frac{-x}{1-e^{-2}}}$
• Oct 18th 2009, 02:16 PM
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.
• Oct 18th 2009, 05:56 PM
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?