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Math Help - Partial fraction

  1. #1
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    Partial fraction

    <br />
\int\frac{x+4}{x^2+5x-6}\,dx<br />

    so factoring the bottom we get

    <br />
\int\frac{x+4}{(x-1)(x+6)}\,dx<br />

    so we have

    A / x-1

    B / x+6

    now what? I know you have to do this

    x+4 = A(x+6) + B(x-1)(x+6)/(x-6)

    now im stuck because of the x-6 in denominator
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  2. #2
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    Quote Originally Posted by ur5pointos2slo View Post

    so we have

    B / x+6

    x+4 = A(x+6) + B(x-1)(x+6)/(x-6)
    You copied it wrong! There should be no denominator by design.
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  3. #3
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    Quote Originally Posted by Thomas154321 View Post
    You copied it wrong! There should be no denominator by design.
    what do you mean "by design"?
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  4. #4
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    ok so Ive tried to work it a little further.

    a= 5/7
    b=2/7

    is this correct?

    then I get

    <br />
1/7 \int\frac{5}{x-1}+\int\frac{2}{x+6}\,dx<br />

    ok now I know this is a du/u but the book has this answer

    1/7 ln|(x+6)^2(x-1)^5| + C

    is there some log property they used to get to this form?
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  5. #5
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    All I meant with "by design" was that the denominator will always cancel because of the way partial fractions work.



    Those values for a and b are correct.

    \frac{1}{7} \int\frac{5}{x-1}+\frac{2}{x+6}~dx

    This is as you say du/u which integrates to ln(u).

    So we have \frac{1}{7}(5ln(x-1)+2ln(x+6)) which is what you have I hope?

    The following are true for logs:
    ln(x) + ln(y) = ln(xy)
    xln(y) = ln(y^x)
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