Question

Integration:

Answer 1

\[\Large \int \frac{xe^x}{(1+x)^2}~dx\]

Answer 2

show us what you did

Answer 3

start from the format of IBP

Answer 4

integration by parts...

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

du = e^x v = ln(1+x) + 1/(1+x)

Answer 5

hmmm

Answer 6

looks like you will need to repeat IBP

Answer 7

I think what I did was probably not the best way..

I had these values instead:
\(u=xe^x\) and \(\Large dv=\frac{dx}{(1+x)^2}\)

\(du=e^x+xe^x dx\) and \(\Large v=-\frac{1}{(1+x)}\)

Answer 8

you can try
u = x e^x du = e^x (x+1) using product rule
dv = 1/(1+x)^2 dx v = -1/(x+1)

use uv - integral(vdu)

Answer 9

I was thinking about how to do this a clever way by seeing that 1/(1+x)^2 is the derivative of the infinite geometric sum of (-x)^n. But I'm still working on it.

Answer 10

correction:
u = x e^x du = e^x (x+1)dx
dv = 1/(1+x)^2 dx v = -1/(x+1)

Answer 11

integral e^x dx-(e^x x)/(x+1)

Answer 12

It may help with this integral to begin with a plain U substitution.
With this choice for U, the integral becomes the following:


thinking of a better way to do this integral... hmm