\(\int\frac{1}{x(x + 1)(x + 2)}dx\) = \(\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2}.....(1)\)
1 = A ( x + 1) (x + 2) + B(x) (x + 2) + c × x(x + 1)
put x = 0, 1 = A(1)(2) ⇒ 1 = 2A ⇒ A = \(\frac{1}{2}\)
put x = -1, 1 = A(0) + B(-1) (-1 + 2) + c(0) ⇒ 1 = -B ⇒ B = -1
put x = -2, 1 = A(0) + B(0) + c(-2), (-2 + 1) ⇒ 1 = 2c ⇒ c = \(\frac{1}{2}\)
∴ \(\int\frac{dx}{x(x + 1)(x + 2)} = \int\frac{\frac{1}{2}}{x} + \frac{-1}{x + 1} + \frac{\frac{1}{2}}{x + 2}dx\)
= \(\frac{1}{2}\)log2 – log(x + 1) + \(\frac{1}{2}\)log(x + 2) +c