`int (x e^(2x))/(1 + 2x)^2 dx` Evaluate the integral

This is integration by parts

rewrite it as:

`int [x e^(2x)] {dx /[(2x + 1)^2]} `

let:
`[x e^(2x)] = u → {e^(2x) + x [2e^(2x)]} dx = du →`
`{e^(2x) + 2x e^(2x)} dx = du → `

factoring out `e^(2x)`

`[e^(2x)](1 + 2x) dx = du `

`dx /[(2x + 1)^2] = dv → `

dividing and multiplying by 2
`(1/2) {2dx /[(2x + 1)^2]} = dv → `
`(1/2) {d(2x + 1) /[(2x + 1)^2]} = dv → `
`(1/2) [(2x + 1)^(-2)] d(2x + 1) = dv → `
`(1/2) [(2x + 1)^(-2+1)]/(-2+1) = v → `
`(1/2) [(2x + 1)^(-1)]/(-1) = v → `

`-1 /[2(2x + 1)] = v `

thus, integrating by parts, you get:

`int u dv = u v - int v du → `

`int [x e^(2x)] {dx /[(2x + 1)^2]} = `

`[x e^(2x)]{-1 /[2(2x + 1)]} - ∫ {-1 /[2(2x + 1)]} [e^(2x)](1 + 2x) dx = `

`(-1/2){[x e^(2x)] /(2x + 1)} + ∫ {[e^(2x)](1 + 2x) /[2(2x + 1)]} dx = `

canceling `(2x + 1)` ,

`(-1/2){[x e^(2x)] /(2x + 1)} + (1/2) ∫ e^(2x) dx = `

`(-1/2){[x e^(2x)] /(2x + 1)} + (1/2) [(1/2)e^(2x)] + C = `

factoring out `[(1/2)e^(2x)]` ,

`[(1/2)e^(2x)] {- [x/(2x + 1)] + (1/2)} + C = `

`[(1/2)e^(2x)] {(- 2x + 2x + 1) /[2(2x + 1)]} + C = `

`[(1/2)e^(2x)] {1 /[2(2x + 1)]} + C `

Thus

`int [x e^(2x)] / [(2x + 1)^2] dx = (1/4) {[e^(2x)] /(2x + 1)} + C`