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Math Help - Evaluate the definite integral ∫1/((x+9)(x62+4)) , x from -3 to 2

  1. #1
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    Evaluate the definite integral ∫1/((x+9)(x62+4)) , x from -3 to 2

    I've been stuck on this question for a while.
    Can someone show me how to do this one, thanks.

    Question is actually

    ∫1/((x+9)(x^2+4))
    Last edited by Kitizhi; March 2nd 2009 at 07:39 PM. Reason: typo
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  2. #2
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    Try partial subsitiuation

    \int{{{1}\over{\left(x+9\right)\,\left(x^2+4\right  )}}}\,dx
    <br />
\int{{{1}\over{\left(x+9\right)\,\left(x^2+4\right  )}}}\,dx<br />
= A/(x+9) + (Bx+C)/(x^2+4)

    solve for A and B and then integrate it will be much easier.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Kitizhi View Post
    I've been stuck on this question for a while.
    Can someone show me how to do this one, thanks.

    Question is actually

    ∫1/((x+9)(x^2+4))
    Partial Fractions:

    \int \frac{1}{(x+9)(x^2+4)} dx

    \frac{A}{(x+9)} +  \frac{Bx+C}{x^2+4}

    Get a common denominator:

    A(x^2 + 4) + (Bx+C)(x+9) * THis is equal to the bottom of your original integral

    So,

    1 = A(x^2 + 4) + (Bx+C)(x+9)

    Now FOIL the right side and collect like terms:
    <br />
= Ax^2 + 4A + Bx^2 + 9Bx + Cx + 9C

    = (A + B)x^2 + (9B +C)x + (4A + 9C)

    Now you can solve for A B & C * Remember that since your numerator in your original integral was 1, you will only put 4A + 9C = 1. You always put your like terms equal to the coefficient of the like term in the numerator of your original integral, so in this case since there are no x or x^2 terms in the numerator of your integral, the other two will be = 0.

    1 = 4A + 9C

    0 = A + B

    0 = 9B + C

    So,

    C = -9B
    A = -B

    then

    1 = 4(-B) + 9(-9B)
    1 = -4B - 81B
    1 = -85B

    B = -\frac{1}{85}

    Then you can solve for the others.

    When you get your A B & C, just plug them back into


    \frac{A}{(x+9)} +  \frac{Bx+C}{x^2+4}

    and this is your new integral!
    Whew, these are usually pretty long and time consuming! Hope this helps!
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  4. #4
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    Thanks alot mollymcf2009,
    I think I can manage from here on
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