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Math Help - Partial fraction with repeated irreducible 2nd-degree polynomial in the denominator

  1. #1
    Junior Member symstar's Avatar
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    Partial fraction with repeated irreducible 2nd-degree polynomial in the denominator

    Woe is me and the partial fractions...

    \int\frac{x-3}{(x^2+2x+4)^2}dx

    I tried working with just the fraction and got the same thing back that I put in, then I tried completing the square and subbing u=x+1 and got the same thing .

    What the heck do I do with this thing?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by symstar View Post
    Woe is me and the partial fractions...

    \int\frac{x-3}{(x^2+2x+4)^2}dx

    I tried working with just the fraction and got the same thing back that I put in, then I tried completing the square and subbing u=x+1 and got the same thing .

    What the heck do I do with this thing?
    \int\frac{x-3}{(x^2+2x+4)^2}\,dx=\int\frac{x-3}{(x^2+2x+1+3)^2}\,dx=\int\frac{x-3}{((x+1)^2+3)^2}\,dx

    Let u=x+1\implies \,du=\,dx

    So we get \int\frac{u-4}{(u^2+3)^2}\,du

    Now we see that we can split it up and get \int\frac{u\,du}{(u^2+3)^2}-4\int\frac{\,du}{(u^2+3)^2}

    For \int\frac{u\,du}{(u^2+3)^2}, let z=u^2+3\implies \,dz=2u\,du, then we get \tfrac{1}{2}\int\frac{\,dz}{z^2}

    For -4\int\frac{\,du}{(u^2+3)^2}, let z=u^2+3\implies \,dz=2u\,du, then we get -2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}}

    let z=3\sec^2\vartheta\implies \,dz=6\sec^2\vartheta\tan\vartheta\,d\vartheta

    Thus, -4\int\frac{\,du}{(u^2+3)^2}=-12\int\frac{(3\sec^2\vartheta-3)(\sec^2\vartheta\tan\vartheta)\,d\vartheta}{9\se  c^4\vartheta\sqrt{3(\sec^2\vartheta-1)}}=-12\int\frac{3\sec^2\vartheta\tan^3\vartheta\,d\var  theta}{9\sqrt{3}\sec^4\vartheta\tan\vartheta} =-\frac{4\sqrt{3}}{9}\int\sin^2\vartheta\,d\vartheta  =-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\vartheta-\tfrac{1}{2}\sin\vartheta\cos\vartheta\right]

    Now, converting back to an equation in z, we see that the anti-derivative is -\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{\sqrt{z-3}}{\sqrt{3}}\right)-\tfrac{1}{2}\frac{\sqrt{3z-9}}{z}\right]

    The anti-derivative (as a whole) is -\frac{1}{2z}-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{\sqrt{z-3}}{\sqrt{3}}\right)-\tfrac{1}{2}\frac{\sqrt{3z-9}}{z}\right]+C

    Since z=u^2+3, then we get -\frac{1}{2(u^2+3)}-\frac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{u\sqrt{3}}{3}\right)-\tfrac{1}{2}\frac{\sqrt{3}u}{u^2+3}\right]+C

    Now, since u=x+1 we *finally* see that \int\frac{(x-3)\,dx}{(x^2+2x+4)^2}=\frac{-1}{2x^2+4x+8}-\tfrac{4\sqrt{3}}{9}\left[\tfrac{1}{2}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)-\tfrac{1}{2}\frac{\sqrt{3}(x+1)}{x^2+2x+4}\right]+C

    I (most likely) made a mistake somewhere in this calculation, but the answer is supposed to be -\frac{4x+7}{6(x^2+2x+4)}-\frac{2\sqrt{3}}{9}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)+C

    Try to make sense of what I did......

    --Chris
    Last edited by Chris L T521; September 14th 2008 at 11:01 PM.
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  3. #3
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    An easier way to find the integral \int\frac{du}{(u^2+3)^2} is simply to make the substitution u=\sqrt{3}\tan{z}. Then du=\sqrt{3}\sec^2{z}\,dz, and
    \int\frac{du}{(u^2+3)^2}=\int\frac{\sqrt{3}\sec^2{  z}\,dz}{(3\tan^2{z}+3)^2}
    \;\;=\frac{1}{3\sqrt{3}}\int\frac{\sec^2{z}}{(\tan  ^2{z}+1)^2}\,dz
    \;\;=\frac{1}{3\sqrt{3}}\int\frac{\sec^2{z}}{(\sec  ^2{z})^2}\,dz
    \;\;=\frac{1}{3\sqrt{3}}\int\frac{1}{\sec^2{z}}\,d  z
    \;\;=\frac{1}{3\sqrt{3}}\int\cos^2{z}\,dz
    Proceeding from there should be much easier.

    --Kevin C.
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  4. #4
    Junior Member symstar's Avatar
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    Wow. I follow most of it... but, -2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}} did you rationalize the numerator here?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by symstar View Post
    Wow. I follow most of it... but, -2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}} did you rationalize the numerator here?
    I'd follow TwistedOne151's way of integrating -4\int\frac{\,du}{u^2+3}, but the way I did it was follows:

    I saw that if z=u^2+3, then \,dz=2u\,du

    That results in -2\int\frac{u\,dz}{z^2}

    Now, z=u^2+3\implies u=\sqrt{z-3}

    So, we now get -2\int\frac{\sqrt{z-3}\,dz}{z^2}

    Oh...I just noticed now that we could just make the substitution z=3\sec^2\vartheta right here.

    But to answer your question, I saw that \sqrt{z-3}=\frac{z-3}{\sqrt{z-3}}, so then the integral became -2\int\frac{(z-3)\,dz}{z^2\sqrt{z-3}}

    As a side note, did anyone catch my little mistake? My answer simplifies to -\frac{4x+1}{6(x^2+2x+4)}-\frac{2\sqrt{3}}{9}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right), but the answer is -\frac{4x+{\color{red}7}}{6(x^2+2x+4)}-\frac{2\sqrt{3}}{9}\tan^{-1}\left(\frac{(x+1)\sqrt{3}}{3}\right)

    I probably shouldn't be doing these problems when I'm dead tired

    --Chris
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  6. #6
    Moo
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    Following Chris' will... lol

    Quote Originally Posted by Chris L T521 View Post
    For -4\int\frac{\,du}{(u^2+3)^2}, let z=u^2+3\implies \,dz=2u\,du, then we get -2\int\frac{u\,dz}{z^2}=-2\int\frac{z-3\,dz}{z^2\sqrt{z-3}}
    Wow my eyes are all blurry

    I think the mistake is here.

    In order not to do any mistake, I always write what du is (or at least, I do it to make people understand ): du=\frac{dz}{2u}

    So -4 \int \frac{du}{(u^2+3)^2} transforms into -2 \int \frac{dz}{{\color{red}u} z^2}

    I haven't checked what followed..
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Moo View Post
    Wow my eyes are all blurry

    I think the mistake is here.

    In order not to do any mistake, I always write what du is (or at least, I do it to make people understand ): du=\frac{dz}{2u}

    So -4 \int \frac{du}{(u^2+3)^2} transforms into -2 \int \frac{dz}{{\color{red}u} z^2}

    I haven't checked what followed..
    Wow...that was....erm...yea

    I think that is the only mistake

    --Chris
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