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Math Help - [SOLVED] Partial fractions integral

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Partial fractions integral

    I'm betting I have to solve it using partial fractions : \int \frac{2x^3+3x^2+2x+4}{(x^2+1)^2}dx. So I found A, B, C and D and I reached to that the solution must be equal to \int \frac{2x+3}{x^2+1}+\frac{4}{(x^2+1)^2}dx=\ln |x^2+1|+4\int \frac{dx}{(x^2+1)^2}. And I'm stuck here! This last integral looks quite simple I know, and is very similar to \arctan (x), but I don't know how to solve it. I tried by integration by parts, u-substitution. Should I try for trigonometric substitution? If yes, please start me up making a good choice for the sub. I'm not used to trigo-sub... Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    I'm betting I have to solve it using partial fractions : \int \frac{2x^3+3x^2+2x+4}{(x^2+1)^2}dx. So I found A, B, C and D and I reached to that the solution must be equal to \int \frac{2x+3}{x^2+1}+\frac{4}{(x^2+1)^2}dx=\ln |x^2+1|+4\int \frac{dx}{(x^2+1)^2}. And I'm stuck here! This last integral looks quite simple I know, and is very similar to \arctan (x), but I don't know how to solve it. I tried by integration by parts, u-substitution. Should I try for trigonometric substitution? If yes, please start me up making a good choice for the sub. I'm not used to trigo-sub... Thanks.
    There is the first part and if you scroll down (you don't have to) there is a tad more but not the full solution, for I know that is not what you want.


    \int\frac{dx}{(x^2+1)^2}

    Let x=\tan(\theta)













    dx=\sec^2(\theta)

    so we have

    \int\frac{sec^2(\theta)}{\sec^4(\theta)}~d\theta=\  int\cos^2(\theta)~d\theta
    Last edited by Mathstud28; July 19th 2008 at 02:31 PM.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Sorry mathstud28, I don't understand what you did. I've scrolled down what I wasn't obligated to, but this confuses me. We would have \int \frac{dx}{x^2}, which isn't true, isn't it?
    And yes, you know me well, I prefer not to have the full answer if possible.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Your answer interests me!
    But I just realized I just showed a formula that fit exactly with this integral! Look at one of my earlier post...
    So the integral equals to \frac{x}{2x^2+2}+\frac{\arctan (x)}{2}+C. Thus the final result follows.
    But as I said, I'd like you to help me more about this trig-sub.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Your answer interests me!
    But I just realized I just showed a formula that fit exactly with this integral! Look at one of my earlier post...
    So the integral equals to \frac{x}{2x^2+2}+\frac{\arctan (x)}{2}+C. Thus the final result follows.
    But as I said, I'd like you to help me more about this trig-sub.

    That is the correct answer, here is the trig sub justification.

    \int\frac{dx}{(x^2+1)^2}


    If we let

    x=\tan(\theta)

    \Rightarrow{dx=\sec^2(\theta)}d\theta

    So we would have

    \int\frac{sec^2(\theta)}{(\tan^2(\theta)+1)^2}~d\t  heta

    =\int\frac{\sec^2(\theta)}{(\sec^2(\theta))^2}~d\t  heta

    =\int\frac{d\theta}{\sec^2(\theta)}

    =\int\cos^2(\theta)~d\theta

    =\frac{1}{2}\int\bigg[1+\cos(2\theta)\bigg]~d\theta

    =\frac{1}{2}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]

    =\frac{1}{2}\bigg[\theta+\sin(\theta)\cos(\theta)\bigg]

    \underbrace{=}_{\text{backsub}}\frac{1}{2}\bigg[\arctan(x)+\sin\left(\arctan(x)\right)\cos\left(\a  rctan(x)\right)\bigg]

    =\frac{1}{2}\bigg[\arctan(x)+\frac{x}{\sqrt{1+x^2}}\cdot\frac{1}{\sq  rt{1+x^2}}\bigg]

    =\frac{1}{2}\bigg[\arctan(x)+\frac{x}{x^2+1}\bigg]

    =\frac{\arctan(x)}{2}+\frac{x}{2(x^2+1)}


    That ought to be detailed enough for you .
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  6. #6
    MHF Contributor arbolis's Avatar
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    That ought to be detailed enough for you .
    Yes! Let me analyze this now.
    Note that in your first post you said
    Let
    and in your last :
    .
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Yes! Let me analyze this now.
    Note that in your first you said and in your last : .
    Oops. ..now I see why you might have been confused . Desole.
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