Let $\frac{1}{x^2(x+1)}$ $= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$ Putting x = 0 in (i), we get 1 = B(0 + 1) ⇒ B = 1Putting x = – 1 in (i), we get 1 = C(–1)2 ⇒ C = 1Comparing coefficients of x2 on both sides of (i), we get0 = A + C ⇒ A = – C ⇒ A = –1$\therefore \frac{1}{x^2(x+1)}$ $= -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}$$\therefore \int\limits_{1}^{3} \frac{dx}{x^2(x+1)}$ $= \int\limits_{1}^{3} \left( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \right) dx$$= \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3}$$= \left[ -\frac{1}{x} + \log\left| \frac{x+1}{x} \right| \right]_{1}^{3}$$= \left( -\frac{1}{3} + 1 \right) + \log\frac{4}{3} - \log2$$= \frac{2}{3} + \log\left( \frac{4}{3} \times \frac{1}{2} \right)$$= \frac{2}{3} + \log\frac{2}{3}$