Let $I=\int \frac{3x-1}{(x+2)^2} dx$ Let $\frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}$ ⇒ 3x – 1 = A(x + 2) + B ...(i)Comparing coefficients of x in (i), we get A = 3 Comparing the coefficients of constant terms in (i), we get2A + B = –1 ...(ii)Put A = 3 in (ii), we get 6 + B = –1 ⇒ B = –7$\therefore \frac{3x-1}{(x+2)^2} = \frac{3}{x+2} - \frac{7}{(x+2)^2}$ $\therefore \int \frac{3x-1}{(x+2)^2} dx$ $= 3 \int \frac{dx}{x+2} - 7 \int \frac{dx}{(x+2)^2}$$= 3 \log|x+2| - 7 \frac{(x+2)^{-1}}{-1} + C$$= 3 \log|x+2| + \frac{7}{x+2} + C$