Question #a20af

1 Answer
Apr 2, 2017

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

Explanation:

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

Performing integration by parts, let:

#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#