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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[\int\limits_{}^{}\frac{ xe ^{2x} }{ (1+2x)^2 }\]

  2. anonymous
    • one year ago
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    ok ive gotten to where you are

  3. anonymous
    • one year ago
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    i mean understanding what you did

  4. ganeshie8
    • one year ago
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    small typo, let me fix it quick

  5. ganeshie8
    • one year ago
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    Let : \(u=xe^{2x} \implies du = (1\cdot e^{2x}+x\cdot 2e^{2x})dx = e^{2x}(1+2x)dx\) and \(dv = \dfrac{1}{(1+2x)^2} dx\implies v = -\dfrac{1}{2(1+2x)}\)

  6. ganeshie8
    • one year ago
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    \[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} - \int e^{2x}(1+2x)\cdot \dfrac{-1}{2(1+2x)} dx\\~\\ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\ \end{align}\]

  7. ganeshie8
    • one year ago
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    rest should be easy is there any other function thats as easy as exponential function to integrate ?

  8. anonymous
    • one year ago
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    so i should let u = e^2x and du = e^2

  9. ganeshie8
    • one year ago
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    are you saying you dont know how to integrate \(e^{2x}\) ?

  10. anonymous
    • one year ago
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    yea

  11. anonymous
    • one year ago
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    u =2x, du =2, dv = e , v =e?

  12. ganeshie8
    • one year ago
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    ohkie, its easy, whats the derivative of \(e^{2x}\) ?

  13. anonymous
    • one year ago
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    e^(2x)?

  14. ganeshie8
    • one year ago
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    heard of chain rule before ?

  15. anonymous
    • one year ago
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    oh

  16. anonymous
    • one year ago
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    e^(2x)2

  17. ganeshie8
    • one year ago
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    right, so an antiderivative of e^(2x)2 is e^(2x)

  18. ganeshie8
    • one year ago
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    another name for "antiderivative" is "indefinite integral"

  19. ganeshie8
    • one year ago
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    lets do few more examples so that you see how to to guess these

  20. ganeshie8
    • one year ago
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    whats the derivative of \(\sin x\) ?

  21. anonymous
    • one year ago
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    cos x

  22. ganeshie8
    • one year ago
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    so an antiderivative of \(\cos x\) is \(\sin x\) : \(\int \cos x\,dx = \sin x+C\)

  23. anonymous
    • one year ago
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    -sinx +c?

  24. ganeshie8
    • one year ago
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    its +sinx because the derivative of \(\sin x\) is \(\cos x\), the antiderivative of \(\cos x\) is \(\sin x+C\)

  25. anonymous
    • one year ago
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    ok got it

  26. ganeshie8
    • one year ago
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    lets do couple more find an antiderivative of \(\large x^2\)

  27. anonymous
    • one year ago
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    x^3/3

  28. ganeshie8
    • one year ago
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    correct. but why ?

  29. anonymous
    • one year ago
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    the formula is x^n+1/n+1 ?

  30. ganeshie8
    • one year ago
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    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

  31. anonymous
    • one year ago
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    ok

  32. ganeshie8
    • one year ago
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    find an antiderivative of \(e^{2x}\)

  33. anonymous
    • one year ago
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    that makes sense ill just from now on relate that

  34. ganeshie8
    • one year ago
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    in other words, what function do u need to differentiate in order to produce \(e^{2x}\)

  35. anonymous
    • one year ago
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    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.

  36. anonymous
    • one year ago
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    i cant do it that way though thats cool

  37. anonymous
    • one year ago
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    (e^(2x))/2

  38. anonymous
    • one year ago
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    also wio that is the correct answer

  39. anonymous
    • one year ago
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    v=e^2/2

  40. ganeshie8
    • one year ago
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    Nice! as you can see integration is indeed a guessing game and you can only get good at it by practice

  41. ganeshie8
    • one year ago
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    are you saying (e^(2x))/2 is an antiderivative of e^(2x) ?

  42. anonymous
    • one year ago
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    yes

  43. ganeshie8
    • one year ago
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    why

  44. anonymous
    • one year ago
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    cause the derivative of that is e^2x

  45. ganeshie8
    • one year ago
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    Perfect! i wanted to hear that from you :)

  46. ganeshie8
    • one year ago
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    so you can work the rest of the problem

  47. ganeshie8
    • one year ago
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    |dw:1440738169733:dw|

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

  49. ganeshie8
    • one year ago
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    Easy, you just need to plugin \(\int e^{ex}\, dx\)

  50. ganeshie8
    • one year ago
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    Easy, you just need to plugin \(\int e^{2x}\, dx\)

  51. ganeshie8
    • one year ago
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    maybe let me just show u

  52. ganeshie8
    • one year ago
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    \[\begin{align}\int\dfrac{ xe ^{2x} }{ (1+2x)^2 }~~ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} - \int e^{2x}(1+2x)\cdot \dfrac{-1}{2(1+2x)} dx\\~\\ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \int \dfrac{e^{2x}}{2}dx \\~\\ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \frac{1}{2} \int e^{2x}\, dx \\~\\ &=~~ xe^{2x}\cdot \dfrac{-1}{2(1+2x)} + \frac{1}{2} \dfrac{e^{2x}}{2} + C \\~\\ \end{align}\]

  53. ganeshie8
    • one year ago
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    we're done! simplify if you want to

  54. anonymous
    • one year ago
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    the answer is supposed to by e^2x/4(1+2x) + C

  55. ganeshie8
    • one year ago
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    you will get that if u simplify

  56. ganeshie8
    • one year ago
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    check this when you have time https://www.khanacademy.org/math/integral-calculus/integration-techniques/integration_by_parts/v/deriving-integration-by-parts-formula

  57. anonymous
    • one year ago
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    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

  58. anonymous
    • one year ago
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    I have been working on that problem all literally since like noon to 10 :25 pm you are my new hero

  59. anonymous
    • one year ago
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    of course i had completely forgotten about the chain rule for e^2x

  60. anonymous
    • one year ago
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    the only thing im really still confused over is how you get -1/2(1+2x) for the first v value?

  61. ganeshie8
    • one year ago
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    differentiate -1/2(1+2x) what do you get ?

  62. ganeshie8
    • one year ago
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    integration takes some time to get used to... because it requirs some guessing and not so straightforward as differentiation

  63. anonymous
    • one year ago
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    does moving the denominator up make it easier? like (1+2x)^-2

  64. ganeshie8
    • one year ago
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    |dw:1440739783369:dw|

  65. anonymous
    • one year ago
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    oh got it

  66. anonymous
    • one year ago
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    i did integral of (1+2x)^-2 = (1+2x)-1/-1

  67. anonymous
    • one year ago
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    you have been a huge huge help

  68. anonymous
    • one year ago
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    thank you so much for spending so much time with me

  69. ganeshie8
    • one year ago
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    looks good! make sure you watch that video from khan academy, it has a very good introduction to integration by parts...

  70. anonymous
    • one year ago
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    ok im gonna watch it now. thank you again for you help and your time

  71. anonymous
    • one year ago
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    \[\int\limits_{?}^{?}\frac{ xe ^{2x}}{ (1+2x)^{2} }; Let U=xe ^{2x}; du=e ^{2x}(1+2x); dv= \frac{ 1 }{ (1+2x)^{2}}; v= -\frac{ 1 }{ 1+2x } \]

  72. anonymous
    • one year ago
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    \[uv-\int\limits_{?}^{?}vdu = -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits_{}^{?}-\frac{ 1 }{ 2(1+2x) }*e ^{2x}(1+2x)*dx\]

  73. anonymous
    • one year ago
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    \[= -\frac{ xe ^{2x} }{ 2(1+2x) }-\int\limits -\frac{ 1 }{ 2 }*e ^{2x}dx = -\frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ 1 }{ 2 }\int\limits e ^{2x}dx\]

  74. anonymous
    • one year ago
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    \[-\frac{ xe ^{2x} }{ 2(1+2x) } + \frac{ 1 }{ 2 }*\frac{ e ^{2x} }{ 2 }= -\frac{ xe ^{2x} }{ 2(1+2x)} + \frac{ e ^{2x} }{ 4 } =-\frac{ 4ex ^{2x} }{ (8+16x) }+\frac{ 2e ^{2x} +4ex ^{2x}}{ (8+16x) }\]

  75. anonymous
    • one year ago
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    \[=\frac{ 2e ^{2x} }{ 8(1+2x) } = \frac{ e ^{2x} }{ 4(1+2x) }\]

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