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Thread: how to integrate -2/(x(x + 2))

  1. #1
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    how to integrate -2/(x(x + 2))

    how would you go about integrating $\displaystyle \int \frac{-2}{x(x+2)} dx$?

    i dont even know where to start?
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  2. #2
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    Quote Originally Posted by renlok View Post
    how would you go about integrating $\displaystyle \int \frac{-2}{x(x+2)} dx$?

    i dont even know where to start?
    Partial Fractions

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

    $\displaystyle A(x+2) + Bx = -2$

    Let $\displaystyle x = 0$ so that $\displaystyle 2A = -2$ hence $\displaystyle A = -1$


    Let $\displaystyle x = -2$ such that $\displaystyle -2B = -2$ and $\displaystyle B = 1$


    The integral therefore becomes $\displaystyle \int \frac{-2}{x(x+2)}\,dx = -\int \frac{1}{x} \,dx+ \int\frac{1}{x+2}\,dx$ which is simpler
    Last edited by e^(i*pi); Feb 14th 2010 at 01:22 PM. Reason: changed partial fraction from 2 to x
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  3. #3
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    Quote Originally Posted by renlok View Post
    how would you go about integrating $\displaystyle \int \frac{-2}{x(x+2)} dx$?

    i dont even know where to start?
    Hi renlok,

    You can write $\displaystyle \frac{-2}{x(x+2)}$ as the sum of two fractions.

    You need to find their numerators.

    $\displaystyle \frac{a}{x}+\frac{b}{x+2}=\frac{-2}{x(x+2)}$

    $\displaystyle \frac{a(x+2)}{x(x+2)}+\frac{bx}{x(x+2)}=\frac{-2}{x(x+2)}$

    $\displaystyle ax+2a+bx=-2+(0)x$

    $\displaystyle 2a=-2,\ a=-1$

    $\displaystyle a+b=0,\ b=-a,\ b=1$

    Now you only need solve

    $\displaystyle \int{\frac{-2}{x(x+2)}}dx=\int{\frac{1}{x+2}}dx-\int{\frac{1}{x}}dx$
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