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

\(\frac{1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}\)

\(1 = A(x+1) + B(x-2)\)

If \(x = -1 \) \(B = \frac{-1}{3}\)

If \(x = 2 \) \(B = \frac{1}{3}\)

\(\int{\frac{dx}{(x-2)(x+1)}} = \int \frac{dx}{3(x-2) - \int \ frac{dx}{3 (x+1)}}\)

\(\int{\frac{dx}{(x-2)(x+1)}} =\frac{1}{3}[ \ln(x-2) - ln(x+1) ]+ C\)

\(\int{\frac{dx}{(x-2)(x+1)}} =\frac{1}{3}[ \ln{ \frac{x-2}{x+1}} ]+ C\)