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Math Help - 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 \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 \int \frac{-2}{x(x+2)} dx?

    i dont even know where to start?
    Partial Fractions

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

    A(x+2) + Bx = -2

    Let x = 0 so that 2A = -2 hence A = -1


    Let x = -2 such that -2B = -2 and B = 1


    The integral therefore becomes \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); February 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 \int \frac{-2}{x(x+2)} dx?

    i dont even know where to start?
    Hi renlok,

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

    You need to find their numerators.

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

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

    ax+2a+bx=-2+(0)x

    2a=-2,\ a=-1

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

    Now you only need solve

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