Question

Question #a20af

Answer 1

`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`