Question

do it
$\bold \wr \implies { \int \: 1 \div x(x + 1)(x + 2)}$
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Answer 1

$\huge\mathcal\colorbox{lavender}{{\color{b}{ꪗour Answer★࿐}}}$

$\int \: \frac{1}{x(x + 1)(x + 2)} dx \\ \frac{1}{x(x + 1)(x + 2)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x + 2} \\ 1 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)...(i) \\ put \: \: x = 0 \: \: in \: \: eqn \: \: (i) \\ 1 = 2A\implies A = \frac{1}{2} \\ put \: \: x = - 1 \\ 1 = B( - 1)( - 1 + 2) \\ 1 = - B \\ B = - 1 \\ put \: \: x = - 2 \\ 1 = C( - 2)( - 2 + 1) \\ C = \frac{1}{2} \\ \frac{1}{x(x + 1)(x + 2)} = \frac{ \frac{1}{2} }{x} + \frac{ - 1}{x + 1} + \frac{ \frac{1}{2} }{x + 2} \\ \int \frac{1}{x(x + 1)(x + 2)} = \frac{1}{2} \int \frac{1}{x} dx - \int \frac{1}{x + 1} dx \: + \int \: \frac{1}{x + 2} dx \\ = \frac{1}{2} log |x| - log |x + 1| + log |x + 2| + c$