Question

How do you integrate `int (xe^x)/((x+1)^2)dx` using integration by parts?

Answer 1

You may try several other choices (I did) before you try `u=xe^x` and `dv = (x+1)^-2 dx`

Explanation:

Following through on the choices above, we find that

`du = (e^x+xe^x) dx` and `v= - 1/(x+1)`

`int (xe^x)/((x+1)^2)dx = -(xe^x)/(x+1)+ int e^x`

`= e^x/(x+1) +C`