int xe^(2x)/(2x+1)^2 dx Find the indefinite integral

Given to solve,
int xe^(2x)/(2x+1)^2 dx
let u= xe^(2x) so , u' = e^(2x) +2xe^(2x)
and
v'= (1/((2x+1)^2)) = (2x+1)^(-2)
v= int (2x+1)^(-2) dx
let t= 2x+1 => dt = 2dx
so v' =t^(-2)
=>  v= int t^(-2) dt /2
        = t^(-2+1) /(-2+1) *(1/2)
so v = (2x+1)^(-2+1) /(-2+1) * (1/2)
 = -1/(2(2x+1))
by applying the integration by parts we get ,
int uv' = uv - int u'v
so,
int xe^(2x)/(2x+1)^2 dx
= (xe^(2x))(-1/(2(2x+1))) - int (e^(2x) +2xe^(2x))(-1/(2(2x+1))) dx
= (xe^(2x))(-1/(2(2x+1))) + int ((e^(2x) +2xe^(2x))/(2(2x+1))) dx
=(xe^(2x))(-1/(2(2x+1))) +(1/2) int (e^(2x) (1+2x)/((2x+1))) dx
=(xe^(2x))(-1/(2(2x+1))) +(1/2) int (e^(2x)) dx
as we know int e^(ax) dx = e^(ax) /a
so,
=-(xe^(2x))(1/(2(2x+1))) +(1/2) (e^(2x))/2
= (-xe^(2x))/(2(2x+1)) +e^(2x)/4 +c