Question

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

Answer 1

sure

Answer 2

you gota perform some log dances here, show me your working out

Answer 3

I'm supposed to do it with integration by parts. I started off my considering (x*e^2x)=U and 1/(1+2x)^2 =dv but I'm not really getting anywhere with this

Answer 4

where are you from btw? @ uni atm?

Answer 5

Yeah I'm at Uni, this is from Calculus II. Finals are over but I'm trying to gain complete mastery before I move on.

Answer 6

i'm in 11th grade

Answer 7

anyway integral..hmm

Answer 8

i'll write it out on paper first if that's okay.. :P a bit rusty

Answer 9

What is Uni?

Answer 10

@Pquevedo2 , \[\int\limits_{}^{} {{xe^{2x}dx} \over {(1+2x)^2}}\]
Is this your question??

Answer 11

that is my question, yes.

Answer 12

I would begin with a u substitution (u=1+2x). That will simplify your denominator considerably, and break your integral into two pieces. Each piece should be solvable by integration by parts.

Answer 13

Basically you will need to solve the INTEGRAL of e^u / u^2. You can do this by integration by parts, where you integrate the (1/u^2) piece and differentiate the (e^u) piece.

Answer 14

rewrite it as:
\[\int\limits [xe^(2x)]{dx/[1+2x^2]}\]
let:
[x e^(2x)] = u → {e^(2x) + x [2e^(2x)]} dx = du →
{e^(2x) + 2x e^(2x)} dx = du →
factoring out e^(2x)

continuing...

Answer 15

i had a tech glitch with the equation thingy lol

Answer 16

[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

Answer 17

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 18

medal please

Answer 19

I would take t= 2x

\[(1/4) \int\limits_{}^{}{ {te^{t} dt} \over {(1+t)}^2}\]
Then add +1 and -1
\[(1/4) \int\limits_{}^{}{ {(t+1-1)e^{t} dt} \over {(1+t)}^2}\]
\[(1/4) \int\limits_{}^{}{ {(t+1)e^{t} dt} \over {(1+t)}^2}- (1/4) \int\limits_{}^{}{ {e^{t} dt} \over {(1+t)}^2}\]
\[(1/4) \int\limits_{}^{}{ {e^{t} dt} \over {(1+t)}}- (1/4) \int\limits_{}^{}{ {e^{t} dt} \over {(1+t)}^2}\]
\[(1/4)\int\limits_{}^{} e^tdt( { {1} \over{ (1+t) } } - { {1} \over {1+t}^2}) \]

Now use this property
\[\int\limits_{}^{} e^xdx( f(x)+f'(x)) = e^x (f(x))+c\]
In our case
\[f(x)= {{1} \over {1+t}}\]
\[f'(x)= {{-1} \over ({1+t})^2}\]

Therefore we get
\[(1/4)e^t ({{1} \over {1+t}})+C \]
Now substitute t = 2x to get
\[(1/4)e^{2x} ({{1} \over {1+2x}})+C \]

Answer 20

wow i need to learn how to do that^

Answer 21

i suck at using the Equation button :P

Answer 22

No problem @Tai . What finally matters is the answer . :D Medal to you @Tai

Answer 23

likewise!

Answer 24

no medal for me :'(

Answer 25

@tai
Please explain these last three steps:

[(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

How did you add the 1/2 to the - [x/(2x + 1)]?
Thanks in advance