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Math Help - Integration by partial fractions?

  1. #1
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    Integration by partial fractions?

    how do you solve this?:

    <br /> <br />
\int \frac {x^3-6}{x^4+6x^2+8} dx<br /> <br />
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  2. #2
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    The first step you'll need to do is to factorise the denominator. Judging by that polynomial, you will end up with two irreducible quadratic factors. I'll give you a hint on a quick way of factorising: there is a 'hidden quadratic'. Substitute a for x^2 and get two factors in terms of a, then convert back to factors in terms of x
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  3. #3
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    Consider

     \frac {x^3-6}{x^4+6x^2+8} = \frac {x^3-6}{(x^2+4)(x^2+2)}
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Fausst View Post
    how do you solve this?:

    <br /> <br />
\int \frac {x^3-6}{x^4+6x^2+8} dx<br /> <br />
    Since x^4+6x^2+8=(x^2+2)(x^2+4), that fraction decomposes to:

    \frac{x^3-6}{x^4+6x^2+8}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+  2}

    Solve for A,B,C,D and then integrate the two pieces separately. To do this, break up \int\frac{Ax+B}{x^2+4}\,dx (for whatever values of A and B you get) into \frac{A}{2}\int\frac{2x}{x^2+4}\,dx+\int\frac{B}{x  ^2+4}\,dx= \frac{A}{2}\ln(x^2+4)+\frac{1}{4}\int\frac{B}{(x/2)^2+1}\,dx

    The latter integral will integrate to something involving \arctan, but I leave the actual calculations to you.
    Last edited by redsoxfan325; October 22nd 2009 at 09:02 AM. Reason: Fixed a typo - I had \sqrt{2} instead of 2
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  5. #5
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    Looking at this again, the first thing you should do is some substitution to get that cubic out of the numerator.

    Let a = x^4 + 6x^2 + 8 then \frac{da}{dx} = 4x^3 + 12x so dx = \frac{da}{4(x^3 + 3x)}

    Now you can make the integral be \int \frac{x^3 + 3x}{4(x^3 + 3x)a} da + \int \frac{-3x - 6}{x^4 + 6x^2 + 8}dx
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by calum View Post
    Looking at this again, the first thing you should do is some substitution to get that cubic out of the numerator.

    Let a = x^4 + 6x^2 + 8 then \frac{da}{dx} = 4x^3 + 12x so dx = \frac{da}{4(x^3 + 3x)}

    Now you can make the integral be \int \frac{x^3 + 3x}{4(x^3 + 3x)a} da + \int \frac{-3x - 6}{x^4 + 6x^2 + 8}dx
    That's true, but he still needs to do partial fraction decomposition on the second integral, so this substitution might not save much (if any) time.
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  7. #7
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    Thanks for replying

    Quote Originally Posted by redsoxfan325 View Post

    Solve for A,B,C,D
    I found that: A=2 B=3 C=-1 D=-3

    but i don't understand what you did to separate

    \int\frac{Ax+B}{x^2+4}\,dx<br />


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  8. #8
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Fausst View Post
    Thanks for replying



    I found that: A=2 B=3 C=-1 D=-3

    but i don't understand what you did to separate

    \int\frac{Ax+B}{x^2+4}\,dx<br />


    Remember that the integral of a sum is the sum of the integrals. So

    \int\frac{Ax+B}{x^2+4}\,dx=\int\frac{Ax}{x^2+4}\,d  x+\int\frac{B}{x^2+4}\,dx

    \int\frac{Ax}{x^2+4}\,dx=\int\frac{\frac{A}{2}\cdo  t2x}{x^2+4}\,dx=\frac{A}{2}\int\frac{2x}{x^2+4}\,d  x

    because you can pull constants out in front of integrals.
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  9. #9
    Super Member redsoxfan325's Avatar
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    @ Fausst: I had typo in my original post, which I have fixed.

    Also, the general rule for \arctan integration is \int\frac{c}{x^2+a}\,dx=\frac{c\sqrt{a}}{a}\tan^{-1}\left(\frac{x}{\sqrt{a}}\right), where a and c are arbitrary constants.
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