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

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Answer 1 · 2016-01-28T16:53:23

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

Following through on the choices above, we find that

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

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

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

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