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  1. #1
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    Feb 2008
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    Help

    integrating 1/(x^2 - 4)

    Please?
    Thanks.
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  2. #2
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    Jan 2008
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    Miami, FL
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    If "S" represents integral, then S 1/(x^2 - 4) dx= S 1/[(x-2)(x+2] dx=
    S (1/4)[1/(x-2) - 1/(x+2)] dx = (1/4) (ln|x-2|-ln|x+2|)+C = (1/4)ln|(x-2)/(x+2)|+C, C in R
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  3. #3
    Newbie
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    Feb 2008
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    Quote Originally Posted by Silver View Post
    integrating 1/(x^2 - 4)

    Please?
    Thanks.
    \int{{\frac{1}{x^2-4}}}

    x^2 -4 = (x-2)(x+2)

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

    Expand:

    \int{\frac{1}{4(x-2)} - \frac{1}{4(x+2)}}

    This is easy now:

    \frac{\ln{|x-2|}}{4} - \frac{\ln{|x+2|}}{4}

    = \frac{\ln\frac{|x-2|}{|x+2|}}{4} + C
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