Let $I = \frac{5x}{(x+1)(x^2+9)}dx$Let $\frac{5x}{(x+1)(x^2+9)}$ $= \frac{A}{x+1} + \frac{Bx+C}{x^2+9}$$\Rightarrow 5x = A(x^2+9) + (Bx+C)(x+1)$ ... (i)Let $x+1=0 \Rightarrow x=-1$∴ 5(-1) = A(1 + 9) ⇒ A $= -\frac{1}{2}$Comparing coefficient of $x^2$ in (i), we get $0 = A + B \Rightarrow B = \frac{1}{2}$Let $x = 0$ $\Rightarrow 0 = 9A + C$ $\Rightarrow C = \frac{9}{2}$$\therefore I = \int \frac{-\frac{1}{2}}{x+1}dx + \int \frac{\frac{1}{2}x + \frac{9}{2}}{x^2+9}dx$