Question #a20af

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Answer 1 · 2017-04-02T18:15:42

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

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

$u=e^x" "=>" "du=e^xcolor(white).dx$

$dv=x/(x+1)^2dx" "=>" "v=lnabs(x+1)+1/(x+1)$

Note that you can find $intx/(x+1)^2dx$ by letting $t=x+1$ , showing that $intx/(x+1)^2dx=int(t-1)/t^2dt=lnabst+1/t$ .

Then:

$I=e^xlnabs(x+1)+e^x/(x+1)-inte^xlnabs(x+1)dx-inte^x/(x+1)dx$

Now perform integration by parts on $inte^x/(x+1)dx$ . Let:

$u=e^x" "=>" "du=e^xcolor(white).dx$

$dv=1/(x+1)dx" "=>" "v=lnabs(x+1)$

So:

$I=e^xlnabs(x+1)+e^x/(x+1)-inte^xlnabs(x+1)dx-(e^xlnabs(x+1)-inte^xlnabs(x+1)dx)$

Many of these terms cancel, leaving just:

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

Attaching the constant of integration:

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