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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{ xe ^{2x} }{ (1+2x)^2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok ive gotten to where you are

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean understanding what you did

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6small typo, let me fix it quick

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Let : \(u=xe^{2x} \implies du = (1\cdot e^{2x}+x\cdot 2e^{2x})dx = e^{2x}(1+2x)dx\) and \(dv = \dfrac{1}{(1+2x)^2} dx\implies v = \dfrac{1}{2(1+2x)}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6\[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{1}{2(1+2x)}  \int e^{2x}(1+2x)\cdot \dfrac{1}{2(1+2x)} dx\\~\\ &=~~ xe^{2x}\cdot \dfrac{1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\ \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6rest should be easy is there any other function thats as easy as exponential function to integrate ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i should let u = e^2x and du = e^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6are you saying you dont know how to integrate \(e^{2x}\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u =2x, du =2, dv = e , v =e?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6ohkie, its easy, whats the derivative of \(e^{2x}\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6heard of chain rule before ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6right, so an antiderivative of e^(2x)2 is e^(2x)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6another name for "antiderivative" is "indefinite integral"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6lets do few more examples so that you see how to to guess these

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6whats the derivative of \(\sin x\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6so an antiderivative of \(\cos x\) is \(\sin x\) : \(\int \cos x\,dx = \sin x+C\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6its +sinx because the derivative of \(\sin x\) is \(\cos x\), the antiderivative of \(\cos x\) is \(\sin x+C\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6lets do couple more find an antiderivative of \(\large x^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the formula is x^n+1/n+1 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6nope, 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6find an antiderivative of \(e^{2x}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that makes sense ill just from now on relate that

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6in other words, what function do u need to differentiate in order to produce \(e^{2x}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was considering how: \[ \left(\frac{u}{v}\right)' = \frac{u'vuv'}{v^2} \]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
 one year ago
Best ResponseYou've already chosen the best response.0i cant do it that way though thats cool

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also wio that is the correct answer

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Nice! as you can see integration is indeed a guessing game and you can only get good at it by practice

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6are you saying (e^(2x))/2 is an antiderivative of e^(2x) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cause the derivative of that is e^2x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Perfect! i wanted to hear that from you :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6so you can work the rest of the problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6dw:1440738169733:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.6Easy, you just need to plugin \(\int e^{ex}\, dx\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6Easy, you just need to plugin \(\int e^{2x}\, dx\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6maybe let me just show u

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6\[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{1}{2(1+2x)}  \int e^{2x}(1+2x)\cdot \dfrac{1}{2(1+2x)} dx\\~\\ &=~~ xe^{2x}\cdot \dfrac{1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\ &=~~ xe^{2x}\cdot \dfrac{1}{2(1+2x)} + \frac{1}{2} \int e^{2x}\, dx \\~\\ &=~~ xe^{2x}\cdot \dfrac{1}{2(1+2x)} + \frac{1}{2} \dfrac{e^{2x}}{2} + C \\~\\ \end{align}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6we're done! simplify if you want to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer is supposed to by e^2x/4(1+2x) + C

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6you will get that if u simplify

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6check this when you have time https://www.khanacademy.org/math/integralcalculus/integrationtechniques/integration_by_parts/v/derivingintegrationbypartsformula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0I have been working on that problem all literally since like noon to 10 :25 pm you are my new hero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0of course i had completely forgotten about the chain rule for e^2x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the only thing im really still confused over is how you get 1/2(1+2x) for the first v value?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6differentiate 1/2(1+2x) what do you get ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6integration takes some time to get used to... because it requirs some guessing and not so straightforward as differentiation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does moving the denominator up make it easier? like (1+2x)^2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6dw:1440739783369:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i did integral of (1+2x)^2 = (1+2x)1/1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have been a huge huge help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much for spending so much time with me

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.6looks good! make sure you watch that video from khan academy, it has a very good introduction to integration by parts...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok im gonna watch it now. thank you again for you help and your time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?}\frac{ xe ^{2x}}{ (1+2x)^{2} }; Let U=xe ^{2x}; du=e ^{2x}(1+2x); dv= \frac{ 1 }{ (1+2x)^{2}}; v= \frac{ 1 }{ 1+2x } \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[uv\int\limits_{?}^{?}vdu = \frac{ xe ^{2x} }{ 2(1+2x) }\int\limits_{}^{?}\frac{ 1 }{ 2(1+2x) }*e ^{2x}(1+2x)*dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[= \frac{ xe ^{2x} }{ 2(1+2x) }\int\limits \frac{ 1 }{ 2 }*e ^{2x}dx = \frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ 1 }{ 2 }\int\limits e ^{2x}dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ xe ^{2x} }{ 2(1+2x) } + \frac{ 1 }{ 2 }*\frac{ e ^{2x} }{ 2 }= \frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ e ^{2x} }{ 4 } =\frac{ 4ex ^{2x} }{ (8+16x) }+\frac{ 2e ^{2x} +4ex ^{2x}}{ (8+16x) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[=\frac{ 2e ^{2x} }{ 8(1+2x) } = \frac{ e ^{2x} }{ 4(1+2x) }\]
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