How do you integrate $\int \frac{x e^{x}}{{(x + 1)}^{2}} d x$ $int (xe^x)/((x+1)^2)dx$ 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 $u = x e^{x}$ $u=xe^x$ and $d v = {(x + 1)}^{- 2} d x$ $dv = (x+1)^-2 dx$

Following through on the choices above, we find that

$d u = (e^{x} + x e^{x}) d x$ $du = (e^x+xe^x) dx$ and $v = - \frac{1}{x + 1}$ $v= - 1/(x+1)$

$\int \frac{x e^{x}}{{(x + 1)}^{2}} d x = - \frac{x e^{x}}{x + 1} + \int e^{x}$ $int (xe^x)/((x+1)^2)dx = -(xe^x)/(x+1)+ int e^x$

$= \frac{e^{x}}{x + 1} + C$ $= e^x/(x+1) +C$

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