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Math Help - Integration

  1. #1
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    Integration

    I cant figure out how to integrate this one.

    integral (x+2)/((x+3)^2 - 4 ) dx

    The ans is
    1/4ln(x+5) +1/12 ln(x+1) +c
    Last edited by heatly; September 27th 2010 at 11:05 PM.
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  2. #2
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    It it \int{\frac{x+2}{(x+3)^2 - 4}\,dx} or \int{\frac{x+2}{(x+3)^2}-4\,dx}?
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  3. #3
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    It the 1st one .
    thanks prove it

    \int{\frac{x+2}{(x+3)^2 - 4}\,dx}
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  4. #4
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    You can solve this using partial fractions.

    \frac{x + 2}{(x + 3)^2 - 4} = \frac{x + 2}{(x + 3 - 2)(x + 3 + 2)}

     = \frac{x + 2}{(x + 1)(x + 5)}.


    Applying partial fractions:

    \frac{A}{x + 1} + \frac{B}{x + 5} = \frac{x + 2}{(x + 1)(x + 5)}

    \frac{A(x + 5) + B(x + 1)}{(x + 1)(x + 5)} = \frac{x + 2}{(x + 1)(x + 5)}

    A(x + 5) + B(x + 1) = x + 2

    Ax + 5A + Bx + B = x + 2

    (A + B)x + 5A+B = 1x + 2.


    Therefore A+B= 1 and 5A+B = 2.

    Solving these simultaneously gives A = \frac{1}{4} and B = \frac{3}{4}.


    Therefore

    \frac{x + 2}{(x + 3)^2 - 4} = \frac{1}{4(x + 1)} + \frac{3}{4(x + 5)}.


    So \int{\frac{x + 2}{(x + 3)^2 - 4}\,dx} = \int{\frac{1}{4(x + 1)} + \frac{3}{4(x + 5)}\,dx}

     = \frac{1}{4}\int{\frac{1}{x + 1}\,dx} + \frac{3}{4}\int{\frac{1}{x + 5}\,dx}

     = \frac{1}{4}\ln{|x + 1|} + \frac{3}{4}\ln{|x + 5|} + C.
    Last edited by Prove It; September 28th 2010 at 04:48 AM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Prove It View Post
    You can solve this using partial fractions.

    \frac{x + 2}{(x + 3)^2 - 4} = \frac{x + 2}{(x + 3 - 2)(x + 3 + 2)}

     = \frac{x + 3}{(x + 1)(x + 5)}.

    ...
    Very interesting the 'identity' \frac{x + 2}{(x + 3 - 2)(x + 3 + 2)}= \frac{x + 3}{(x + 1)(x + 5)}...from which it derives that x+2=x+3...

    Kind regards

    \chi \sigma
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  6. #6
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    I do hate when typos go all the way through the post... Fixed now...
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