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

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Given to solve,

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

let `u= xe^(2x) so , u' = e^(2x) +2xe^(2x)`


`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`


`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`


=`-(xe^(2x))(1/(2(2x+1))) +(1/2) (e^(2x))/2 `

= `(-xe^(2x))/(2(2x+1)) +e^(2x)/4 +c`

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