Question

PLZ HELP
integrate (x*e^(2x))/(2x+1)^2

Answer 1

i tried using parts but it just makes the problem worse

Answer 2

I'm tempted to try u = 2x+1 - simple substitution.

Answer 3

Maybe \[u = \frac{1}{2x+1}\] That might have some promise.

Answer 4

math processing error?

Answer 5

With what? I've never had a pencil give that sort of error.

Answer 6

Maybe [Math Processing Error] That might have some promise.

Answer 7

i even tried using partial fractions

Answer 8

Seriously, u = 2x+1 is a good way to go. Let's see what you get after the substitution.

Answer 9

that doesn't give me any result

Answer 10

Why not? I get \[\frac{1}{4e}\int \frac{e^{u}}{u}-\frac{e^{u}}{u^{2}}\;du\]

Answer 11

∫ [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

∫ u dv = u v - ∫ v du →

∫ [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

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

Answer 12

That works, too. Good work. There's not usually only one way to proceed. Feel free to try stuff.