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Math Help - Fraction integral need help!

  1. #1
    Junior Member
    Joined
    Nov 2008
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    30

    Fraction integral need help!

    \int\frac{3x}{(x-1)(x^2+6)}

    I got this far-

    \int\frac{3x}{(x-1)(x^2+6)} = \frac{A}{x-1}+\frac{Bx+C}{x^2+6}

    But after that I don't know how to solve for it because of the Bx+C

    Any help is appreciated! Test tomorrow and I need to know this...
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  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    Start from
    \frac{3x}{(x-1)(x^2+6)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+6}
    multiply by the common denominator (x-1)(x^2+6):
    3x=A(x^2+6)+(Bx+C)(x-1)
    3x=Ax^2+6A+Bx^2-Bx+Cx-C
    3x=(A+B)x^2+(C-B)x+(6A-C)
    Now equate coefficients.

    --Kevin C.
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