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Thread: 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; Sep 27th 2010 at 11:05 PM.
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  2. #2
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    It it $\displaystyle \int{\frac{x+2}{(x+3)^2 - 4}\,dx}$ or $\displaystyle \int{\frac{x+2}{(x+3)^2}-4\,dx}$?
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    It the 1st one .
    thanks prove it

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

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

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


    Applying partial fractions:

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

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

    $\displaystyle A(x + 5) + B(x + 1) = x + 2$

    $\displaystyle Ax + 5A + Bx + B = x + 2$

    $\displaystyle (A + B)x + 5A+B = 1x + 2$.


    Therefore $\displaystyle A+B= 1$ and $\displaystyle 5A+B = 2$.

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


    Therefore

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


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

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

    $\displaystyle = \frac{1}{4}\ln{|x + 1|} + \frac{3}{4}\ln{|x + 5|} + C$.
    Last edited by Prove It; Sep 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.

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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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