anyone good with integration by parts?
Integrate ((x)e^(2x))/(1+2x)^2

- anonymous

anyone good with integration by parts?
Integrate ((x)e^(2x))/(1+2x)^2

- katieb

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- anonymous

\[\int\limits_{}^{}\frac{ xe ^{2x} }{ (1+2x)^2 }\]

- anonymous

ok ive gotten to where you are

- anonymous

i mean understanding what you did

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## More answers

- ganeshie8

small typo, let me fix it quick

- ganeshie8

Let :
\(u=xe^{2x} \implies du = (1\cdot e^{2x}+x\cdot 2e^{2x})dx = e^{2x}(1+2x)dx\)
and \(dv = \dfrac{1}{(1+2x)^2} dx\implies v = -\dfrac{1}{2(1+2x)}\)

- ganeshie8

\[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} - \int e^{2x}(1+2x)\cdot \dfrac{-1}{2(1+2x)} dx\\~\\
&=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\
\end{align}\]

- ganeshie8

rest should be easy
is there any other function thats as easy as exponential function to integrate ?

- anonymous

so i should let u = e^2x and du = e^2

- ganeshie8

are you saying you dont know how to integrate \(e^{2x}\) ?

- anonymous

yea

- anonymous

u =2x, du =2, dv = e , v =e?

- ganeshie8

ohkie, its easy, whats the derivative of \(e^{2x}\) ?

- anonymous

e^(2x)?

- ganeshie8

heard of chain rule before ?

- anonymous

oh

- anonymous

e^(2x)2

- ganeshie8

right, so an antiderivative of e^(2x)2 is e^(2x)

- ganeshie8

another name for "antiderivative" is "indefinite integral"

- ganeshie8

lets do few more examples
so that you see how to to guess these

- ganeshie8

whats the derivative of \(\sin x\) ?

- anonymous

cos x

- ganeshie8

so an antiderivative of \(\cos x\) is \(\sin x\) :
\(\int \cos x\,dx = \sin x+C\)

- anonymous

-sinx +c?

- ganeshie8

its +sinx
because the derivative of \(\sin x\) is \(\cos x\),
the antiderivative of \(\cos x\) is \(\sin x+C\)

- anonymous

ok got it

- ganeshie8

lets do couple more
find an antiderivative of \(\large x^2\)

- anonymous

x^3/3

- ganeshie8

correct. but why ?

- anonymous

the formula is x^n+1/n+1 ?

- ganeshie8

nope, don't use the formula
use the fact that differentiating x^3/3 gives you x^2,
so x^3/3 is an antiderivative of x^2

- anonymous

ok

- ganeshie8

find an antiderivative of \(e^{2x}\)

- anonymous

that makes sense ill just from now on relate that

- ganeshie8

in other words, what function do u need to differentiate in order to produce \(e^{2x}\)

- anonymous

I was considering how: \[
\left(\frac{u}{v}\right)' = \frac{u'v-uv'}{v^2}
\]I put \(v=1+2x\), and then got the equation: \[
(1+2x)u' - (2)u = xe^{2x}
\]I divide both sides by \(x\) and got: \[
\frac{u'-2u}{x} +2u' = e^{2x}
\]I came to the impression that if I can get \(u'=2u\), I could make things simpler.
So I used \(u=Ce^{2x}\).\[
\frac{2Ce^{2x}-4Ce^{2x}}{x} +4Ce^{2x} = e^{2x} \implies 4Ce^{2x} =e^{2x} \implies C=\frac 14
\]That got me to my solution of\[
\frac uv = \frac{\frac{1}{4}e^{2x}}{1+2x} = \frac{e^{2x}}{4+8x}
\]This isn't a reliable method, but I think it's one of the few cases where you can reverse engineer the quotient rule.

- anonymous

i cant do it that way though thats cool

- anonymous

(e^(2x))/2

- anonymous

also wio that is the correct answer

- anonymous

v=e^2/2

- ganeshie8

Nice! as you can see integration is indeed a guessing game and you can only get good at it by practice

- ganeshie8

are you saying (e^(2x))/2 is an antiderivative of e^(2x) ?

- anonymous

yes

- ganeshie8

why

- anonymous

cause the derivative of that is e^2x

- ganeshie8

Perfect! i wanted to hear that from you :)

- ganeshie8

so you can work the rest of the problem

- ganeshie8

|dw:1440738169733:dw|

- anonymous

i still dont know where to go from where we are though. does it go \[\frac{ -xe ^{2x}}{ 2( 1+2x)}+uv-\int\limits_{}^{}vdu\]

- ganeshie8

Easy, you just need to plugin \(\int e^{ex}\, dx\)

- ganeshie8

Easy, you just need to plugin \(\int e^{2x}\, dx\)

- ganeshie8

maybe let me just show u

- ganeshie8

\[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} - \int e^{2x}(1+2x)\cdot \dfrac{-1}{2(1+2x)} dx\\~\\
&=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\
&=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \frac{1}{2} \int e^{2x}\, dx \\~\\
&=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \frac{1}{2} \dfrac{e^{2x}}{2} + C \\~\\
\end{align}\]

- ganeshie8

we're done! simplify if you want to

- anonymous

the answer is supposed to by e^2x/4(1+2x) + C

- ganeshie8

you will get that if u simplify

- ganeshie8

check this when you have time
https://www.khanacademy.org/math/integral-calculus/integration-techniques/integration_by_parts/v/deriving-integration-by-parts-formula

- anonymous

i simplified and got the answer i needed. THANK YOU SOOO much! i will look that tonight after i study this problem a little more. I am clearly lacking some basic understanding

- anonymous

I have been working on that problem all literally since like noon to 10 :25 pm you are my new hero

- anonymous

of course i had completely forgotten about the chain rule for e^2x

- anonymous

the only thing im really still confused over is how you get -1/2(1+2x) for the first v value?

- ganeshie8

differentiate -1/2(1+2x)
what do you get ?

- ganeshie8

integration takes some time to get used to... because it requirs some guessing and not so straightforward as differentiation

- anonymous

does moving the denominator up make it easier? like (1+2x)^-2

- ganeshie8

|dw:1440739783369:dw|

- anonymous

oh got it

- anonymous

i did integral of (1+2x)^-2 = (1+2x)-1/-1

- anonymous

you have been a huge huge help

- anonymous

thank you so much for spending so much time with me

- ganeshie8

looks good!
make sure you watch that video from khan academy, it has a very good introduction to integration by parts...

- anonymous

ok im gonna watch it now. thank you again for you help and your time

- anonymous

\[\int\limits_{?}^{?}\frac{ xe ^{2x}}{ (1+2x)^{2} }; Let U=xe ^{2x}; du=e ^{2x}(1+2x); dv= \frac{ 1 }{ (1+2x)^{2}}; v= -\frac{ 1 }{ 1+2x } \]

- anonymous

\[uv-\int\limits_{?}^{?}vdu = -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits_{}^{?}-\frac{ 1 }{ 2(1+2x) }*e ^{2x}(1+2x)*dx\]

- anonymous

\[= -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits -\frac{ 1 }{ 2 }*e ^{2x}dx = -\frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ 1 }{ 2 }\int\limits e ^{2x}dx\]

- anonymous

\[-\frac{ xe ^{2x} }{ 2(1+2x) } + \frac{ 1 }{ 2 }*\frac{ e ^{2x} }{ 2 }= -\frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ e ^{2x} }{ 4 } =-\frac{ 4ex ^{2x} }{ (8+16x) }+\frac{ 2e ^{2x} +4ex ^{2x}}{ (8+16x) }\]

- anonymous

\[=\frac{ 2e ^{2x} }{ 8(1+2x) } = \frac{ e ^{2x} }{ 4(1+2x) }\]

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