$\displaystyle \int \frac {1}{x^2-1} dx $ any ideas? cheers
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Originally Posted by jvignacio $\displaystyle \int \frac {1}{x^2-1} dx $ any ideas? cheers $\displaystyle x^2-1=(x-1)(x+1)$ $\displaystyle \frac{1}{x^2-1}=\frac 12 \left(\frac{1}{x-1}-\frac{1}{x+1}\right)$
$\displaystyle \int \frac {1}{x^2-1} dx = -\int \frac {1}{1-x^2} dx$ $\displaystyle = - arctanh (x)$ or $\displaystyle = - arccoth (x)$
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