Question

Integrate the rational function: \(\frac{x}{(x+1)(x+2)}\)

Answer 1

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

x = A(x-1)(x+2)+B(x+2)+C(x-1)²
Substituting x = 1, we obtain
B = \(\frac{1}{3}\)
Equating the coefficients of x2 and constant term, we obtain
A + C = 0
−2A + 2B + C = 0
On solving, we obtain

A = \(\frac{2}{9}\) and C = -\(\frac{2}{9}\)

∴\(\frac{x}{(x+1)^2(x+2)} = \frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}\)

\(\Rightarrow \int \frac{x}{(x+1)^2(x+2)}dx = \frac{2}{9}\int\frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1)^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx\)

=\(\frac{2}{9} \log\mid x-1\mid+\frac{1}{3}\bigg(\frac{-1}{x-1}\bigg)-\frac{2}{9}\log\mid x+2\mid+C\)

=\(\frac{2}{9}\log\mid\frac{x-1}{x+2}\mid-\frac{1}{3(x-1)}+C\)