$\displaystyle \int \frac{xe^x}{(x+1)^2}dx$

i dunno what would i substitute to?

u = e^x

du = e^xdx

dv = (x)/(1+x)^2

v = (x^4-2)/(x+1)^4

Helpp :(

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- Jul 11th 2007, 05:26 AM^_^Engineer_Adam^_^Integration by parts again
$\displaystyle \int \frac{xe^x}{(x+1)^2}dx$

i dunno what would i substitute to?

u = e^x

du = e^xdx

dv = (x)/(1+x)^2

v = (x^4-2)/(x+1)^4

Helpp :( - Jul 11th 2007, 05:33 AMJhevon
- Jul 11th 2007, 05:48 AM^_^Engineer_Adam^_^
thanks for the subs and it worked well :)