Let $I = \int \frac{x}{(x+1)(x+2)} dx$ Let $\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$ $\Rightarrow x = A(x+2) + B(x+1)$ ...(i) Putting $x = -1$ in (i), we get $-1 = A(-1+2)$ $\Rightarrow A = -1$ Putting $x = -2$ in (i), we get $-2 = B(-2+1)$ $\Rightarrow B = 2$$\therefore \frac{x}{(x+1)(x+2)} = \frac{-1}{x+1} + \frac{2}{x+2}$$\therefore I = \int \frac{x}{(x+1)(x+2)} dx$ $= \int \left[ \frac{-1}{x+1} + \frac{2}{x+2} \right] dx$$= \int \frac{-1}{x+1} dx + \int \frac{2}{x+2} dx$ $= -\log\lvert x+1\rvert + 2\log\lvert x+2\rvert + C$$= -\log\lvert x+1\rvert + \log \lvert x+2\rvert^2 + C$$= \log \left\lvert \frac{(x+2)^2}{x+1} \right\rvert + C$