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**Silver** · February 27th 2008, 02:41 PM

integrating 1/(x^2 - 4)

Please?
Thanks.

**Gauss** · February 27th 2008, 02:49 PM

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

**Weierstrass** · February 27th 2008, 02:57 PM

Originally Posted by Silver

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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