Let $I=\int \frac{1}{(x^2+1)(x^2+4)}dx$ Now consider $\frac{1}{(x^2+1)(x^2+4)}$ Put $x^2 = t$ Now let $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$ $\Rightarrow 1 = A(t+4) + B(t+1)$ ...(i)Put $t = -1$ in (i), we get, $1 = A(-1+4)$ $\Rightarrow A = \frac{1}{3}$ Put $t = -4$ in (i), we get, $1 = B(-4+1)$ $\Rightarrow B = -\frac{1}{3}$$\therefore \frac{1}{(t+1)(t+4)}$ $= \frac{1}{3(t+1)} - \frac{1}{3(t+4)}$ $= \frac{1}{3(x^2+1)} - \frac{1}{3(x^2+4)}$Now, $I=\int \left[ \frac{1}{3(x^2+1)} - \frac{1}{3(x^2+4)} \right]dx$