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#
