Let $I=\int\frac{5x}{(x+1)(x^2-4)}dx$ Let $\frac{5x}{(x+1)(x^2-4)}$ $=\frac{5x}{(x+1)(x+2)(x-2)}$ $=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x-2}$ ⇒ 5x = A(x + 2) (x – 2) + B(x + 1) (x – 2) + C (x + 1) (x + 2) ...(i)Putting x = –1 in (i), we get – 5 = A(–1 + 2) (–1 –2) ⇒ A $=\frac{5}{3}$Putting x = –2 in (i), we get – 10 = (–2 + 1)(–2 – 2) ⇒ B $=-\frac{5}{2}$Putting x = 2 in (i), we get 10 = C(2 + 1) (2 + 2) ⇒ C $=\frac{5}{6}$$\therefore \frac{5x}{(x+1)(x^2-4)}$ $=\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}$ $\therefore I=\int\frac{5x}{(x+1)(x^2-4)}dx$ $=\frac{5}{3}\int\frac{dx}{x+1}-\frac{5}{2}\int\frac{dx}{x+2}+\frac{5}{6}\int\frac{dx}{x-2}$$=\frac{5}{3}\log|x+1|-\frac{5}{2}\log|x+2|+\frac{5}{6}\log|x-2|+C$