Question

Help me find the integral of (x*e^2x)/(1+2x)^2

Answer 1

∫ [x e^(2x)] / [(2x + 1)^2] dx =

it's by parts:

rewrite it as:

∫ [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:

∫ 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

thus, in conclusion:

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

Answer 2

wow tai.. nice effort..:):)

Answer 3

medal please ;)

Answer 4

Tai you can't post a question and answer it yourself just to receive a medal :/

Answer 5

it wasn't my question :P

Answer 6

@TuringTest .. it was posted by someone else earlier...

Answer 7

ok, I'm trying to see if it's right...
wolfram seems to think not....

Answer 8

i might have messed up actually lol

Answer 9

no I think it's right actually

Answer 10

turingtest are you a moderator?

Answer 11

Maybe u = 2x + 1, 2x = u - 1 and x = (u-1)/2. This should simplify to\[\frac{1}{2}\int\limits \frac{e^{u-1} \frac{u-1}{2}}{u^2}du\]And I think it's due to the weirdness of Wolfram to write the solution, haha.

Answer 12

meh we'll get there one day :P

Answer 13

dawn of the super-supercomputers

Answer 14

ok now it agrees with Tai but won't explain :C
http://www.wolframalpha.com/input/?i=integral+%28x*e%5E%282x%29%29%2F%281%2B2x%29%5E2dx

Answer 15

Yeah, this is a strange integral. I think or partial fractions or doing by Tai's way. Partial fraction is going to be messy, it seems.