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Math Help - An integral, partial fractions

  1. #1
    MHF Contributor arbolis's Avatar
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    An integral, partial fractions

    This one is a really tough one for me. I'm betting an integration via partial fractions : \int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}dx with a \neq 0.
    I could reach that the integral is worth \int 1+\frac{-a^2x^2+6ax-9+2a^2x-6a}{a^2x^2-6ax+10}dx. I don't know how to proceed from it. A first attempt would be to factorize that denominator, but it's not an easy task for me!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    This one is a really tough one for me. I'm betting an integration via partial fractions : \int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}dx with a \neq 0.
    I could reach that the integral is worth \int 1+\frac{-a^2x^2+6ax-9+2a^2x-6a}{a^2x^2-6ax+10}dx. I don't know how to proceed from it. A first attempt would be to factorize that denominator, but it's not an easy task for me!
    Try this:

    \int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}\,dx=\int\frac{2a^2x-6a}{a^2x^2-6ax+10}\,dx+\int\frac{1}{a^2x^2-6ax+10}\,dx

    The first one can be solved using substitution.

    The second one you'll have to experiment with. I'll work on it now. I'll post something when I get it.

    --Chris
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Here's a hint for the second integral:

    \int\frac{1}{a^2x^2-6ax+10}\,dx=\int\frac{1}{a^2x^2-6ax+9+1}\,dx

    \implies\int\frac{1}{(ax-3)^2+1}\,dx

    This is something you can integrate. Take it from here!

    This one didn't need partial fractions after all...

    --Chris
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Try this:

    \int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}\,dx=\int\frac{2a^2x-6a}{a^2x^2-6ax+10}\,dx+\int\frac{1}{a^2x^2-6ax+10}\,dx

    The first one can be solved using substitution.

    The second one you'll have to experiment with. I'll work on it now. I'll post something when I get it.

    --Chris
    \int\frac{dx}{a^2x^2-6ax+10}

    =\frac{1}{a^2}\frac{dx}{x^2-\frac{6}{a}x+\frac{10}{a^2}}

    =\frac{1}{a^2}\int\frac{dx}{\left(x-\frac{3}{a}\right)^2+\frac{1}{a^2}}

    So firstly let

    z=x-\frac{3}{a}

    So

    dx=dz

    So we have

    \frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}

    Now let z=\frac{1}{a}\tan(\theta)

    so

    dz=\frac{1}{a}\sec^2(\theta)

    So we have

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

    =\frac{1}{a}\int~d\theta

    =\frac{\theta}{a}

    Backsubbing we get

    \int\frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}

    =\frac{1}{a}\arctan\left(za\right)


    So

    \int\frac{dx}{a^2x^2-6ax+10}=\frac{\arctan\left(ax-3\right)}{a}
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int\frac{dx}{a^2x^2-6ax+10}

    =\frac{1}{a^2}\frac{dx}{x^2-\frac{6}{a}x+\frac{10}{a^2}}

    =\frac{1}{a^2}\int\frac{dx}{\left(x-\frac{3}{a}\right)^2+\frac{1}{a^2}}

    So firstly let

    z=x-\frac{3}{a}

    So

    dx=dz

    So we have

    \frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}

    Now let z=\frac{1}{a}\tan(\theta)

    so

    dz=\frac{1}{a}\sec^2(\theta)

    So we have

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

    =\frac{1}{a}\int~d\theta

    =\frac{\theta}{a}

    Backsubbing we get

    \int\frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}

    =\frac{1}{a}\arctan\left(za\right)


    So

    \int\frac{dx}{a^2x^2-6ax+10}=\frac{\arctan\left(ax-3\right)}{a}

    You made it too hard...

    \int\frac{1}{a^2x^2-6ax+10}\,dx=\int\frac{1}{a^2x^2-6ax+9+1}\,dx=\int\frac{1}{(ax-3)^2+1}\,dx =\frac{1}{a}\tan^{-1}(ax-3)+C

    --Chris
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You made it too hard...

    \int\frac{1}{a^2x^2-6ax+10}\,dx=\int\frac{1}{a^2x^2-6ax+9+1}\,dx=\int\frac{1}{(ax-3)^2+1}\,dx =\frac{1}{a}\tan^{-1}(ax-3)+C

    --Chris
    I didn't make anything too hard, I just did it in a different style than you.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I didn't make anything too hard, I just did it in a different style than you.
    Ok. Fine. It wasn't too hard. It was just too long...

    Your technique is very interesting. I like it.

    --Chris
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Ok. Fine. It wasn't too hard. It was just too long...

    --Chris
    I just showed ALL my steps.
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