anonymous
  • anonymous
anyone good with integration by parts? Integrate ((x)e^(2x))/(1+2x)^2
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{}^{}\frac{ xe ^{2x} }{ (1+2x)^2 }\]
anonymous
  • anonymous
ok ive gotten to where you are
anonymous
  • anonymous
i mean understanding what you did

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ganeshie8
  • ganeshie8
small typo, let me fix it quick
ganeshie8
  • 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
  • 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
  • ganeshie8
rest should be easy is there any other function thats as easy as exponential function to integrate ?
anonymous
  • anonymous
so i should let u = e^2x and du = e^2
ganeshie8
  • ganeshie8
are you saying you dont know how to integrate \(e^{2x}\) ?
anonymous
  • anonymous
yea
anonymous
  • anonymous
u =2x, du =2, dv = e , v =e?
ganeshie8
  • ganeshie8
ohkie, its easy, whats the derivative of \(e^{2x}\) ?
anonymous
  • anonymous
e^(2x)?
ganeshie8
  • ganeshie8
heard of chain rule before ?
anonymous
  • anonymous
oh
anonymous
  • anonymous
e^(2x)2
ganeshie8
  • ganeshie8
right, so an antiderivative of e^(2x)2 is e^(2x)
ganeshie8
  • ganeshie8
another name for "antiderivative" is "indefinite integral"
ganeshie8
  • ganeshie8
lets do few more examples so that you see how to to guess these
ganeshie8
  • ganeshie8
whats the derivative of \(\sin x\) ?
anonymous
  • anonymous
cos x
ganeshie8
  • ganeshie8
so an antiderivative of \(\cos x\) is \(\sin x\) : \(\int \cos x\,dx = \sin x+C\)
anonymous
  • anonymous
-sinx +c?
ganeshie8
  • ganeshie8
its +sinx because the derivative of \(\sin x\) is \(\cos x\), the antiderivative of \(\cos x\) is \(\sin x+C\)
anonymous
  • anonymous
ok got it
ganeshie8
  • ganeshie8
lets do couple more find an antiderivative of \(\large x^2\)
anonymous
  • anonymous
x^3/3
ganeshie8
  • ganeshie8
correct. but why ?
anonymous
  • anonymous
the formula is x^n+1/n+1 ?
ganeshie8
  • 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
  • anonymous
ok
ganeshie8
  • ganeshie8
find an antiderivative of \(e^{2x}\)
anonymous
  • anonymous
that makes sense ill just from now on relate that
ganeshie8
  • ganeshie8
in other words, what function do u need to differentiate in order to produce \(e^{2x}\)
anonymous
  • 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
  • anonymous
i cant do it that way though thats cool
anonymous
  • anonymous
(e^(2x))/2
anonymous
  • anonymous
also wio that is the correct answer
anonymous
  • anonymous
v=e^2/2
ganeshie8
  • ganeshie8
Nice! as you can see integration is indeed a guessing game and you can only get good at it by practice
ganeshie8
  • ganeshie8
are you saying (e^(2x))/2 is an antiderivative of e^(2x) ?
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
why
anonymous
  • anonymous
cause the derivative of that is e^2x
ganeshie8
  • ganeshie8
Perfect! i wanted to hear that from you :)
ganeshie8
  • ganeshie8
so you can work the rest of the problem
ganeshie8
  • ganeshie8
|dw:1440738169733:dw|
anonymous
  • 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
  • ganeshie8
Easy, you just need to plugin \(\int e^{ex}\, dx\)
ganeshie8
  • ganeshie8
Easy, you just need to plugin \(\int e^{2x}\, dx\)
ganeshie8
  • ganeshie8
maybe let me just show u
ganeshie8
  • 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
  • ganeshie8
we're done! simplify if you want to
anonymous
  • anonymous
the answer is supposed to by e^2x/4(1+2x) + C
ganeshie8
  • ganeshie8
you will get that if u simplify
ganeshie8
  • 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
  • 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
  • anonymous
I have been working on that problem all literally since like noon to 10 :25 pm you are my new hero
anonymous
  • anonymous
of course i had completely forgotten about the chain rule for e^2x
anonymous
  • anonymous
the only thing im really still confused over is how you get -1/2(1+2x) for the first v value?
ganeshie8
  • ganeshie8
differentiate -1/2(1+2x) what do you get ?
ganeshie8
  • ganeshie8
integration takes some time to get used to... because it requirs some guessing and not so straightforward as differentiation
anonymous
  • anonymous
does moving the denominator up make it easier? like (1+2x)^-2
ganeshie8
  • ganeshie8
|dw:1440739783369:dw|
anonymous
  • anonymous
oh got it
anonymous
  • anonymous
i did integral of (1+2x)^-2 = (1+2x)-1/-1
anonymous
  • anonymous
you have been a huge huge help
anonymous
  • anonymous
thank you so much for spending so much time with me
ganeshie8
  • ganeshie8
looks good! make sure you watch that video from khan academy, it has a very good introduction to integration by parts...
anonymous
  • anonymous
ok im gonna watch it now. thank you again for you help and your time
anonymous
  • 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
  • anonymous
\[uv-\int\limits_{?}^{?}vdu = -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits_{}^{?}-\frac{ 1 }{ 2(1+2x) }*e ^{2x}(1+2x)*dx\]
anonymous
  • 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
  • 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
  • anonymous
\[=\frac{ 2e ^{2x} }{ 8(1+2x) } = \frac{ e ^{2x} }{ 4(1+2x) }\]

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