How do you use partial fraction decomposition to decompose the fraction to integrate ${(x^{2} - 4)}^{- 1}$ $(x^2-4)^-1$ ?

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Answer 1 · 2015-05-23T17:02:54

${(x^{2} - 4)}^{- 1} i s \frac{1}{x^{2} - 4}$ $(x^2-4)^-1 is 1/(x^2-4)$

Its partial fraction would be $\frac{1}{4} [\frac{1}{x - 2} - \frac{1}{x + 2}]$ $1/4[1/(x-2) -1/(x+2)]$

Integration would give $\frac{1}{4} [ln (x - 2) - ln (x + 2)}$ $1/4[ln(x-2) -ln(x+2)}$

= $\frac{1}{4} ln (\frac{x - 2}{x + 2})$ $1/4 ln((x-2)/(x+2))$

How do you use partial fraction decomposition to decompose the fraction to integrate (x2−4)−1(x^2-4)^-1?