Question

What is the integral of `int [ (xe^(2x)) / (2x + 1)^2 ]dx`?

Answer 1

Another method:

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

We can try integration by parts with `u=e^(2x)` and `dv=x/(2x+1)^2dx`.

Note that `v=intx/(2x+1)^2dx`. Letting `t=2x+1`, this implies that `x=1/2(t-1)` and that `dt=2dx=>dx=1/2dt`, so `v=int(1/2(t-1))/t^2 1/2dt=1/4int(1/t-1/t^2)dt=1/4lnabst+1/(4t)...`

Also, `du=2e^(2x)dx`.

Then:

`I=1/4e^(2x)(lnabs(2x+1)+1/(2x+1))-int2e^(2x)1/4(lnabs(2x+1)+1/(2x+1))dx`

`I=e^(2x)/4lnabs(2x+1)+e^(2x)/(4(2x+1))-1/2inte^(2x)lnabs(2x+1)dx-1/2inte^(2x)/(2x+1)dx`

Trying IBP on the second integral, let:

`u=-1/2e^(2x)=>du=-e^(2x)dx`
`dv=dx/(2x+1)=>v=1/2lnabs(2x+1)`

So:

`I=e^(2x)/4lnabs(2x+1)+e^(2x)/(4(2x+1))-1/2inte^(2x)lnabs(2x+1)dx-e^(2x)/4lnabs(2x+1)+1/2inte^(2x)lnabs(2x+1)dx`

`I=e^(2x)/(4(2x+1))+C`