Let $I=\int\frac{(x^2+1)(x^2+2)}{(x^3+3)(x^2+4)}dx$ Let $x^2=t$, then $\frac{(t+1)(t+2)}{(t+3)(t+4)}$ $=\frac{t^2+3t+2}{t^2+7t+12}$ $=\left(1+\frac{-4t-10}{t^2+7t+12}\right)=1-\left[\frac{4t+10}{(t+3)(t+4)}\right]$ Let $\frac{4t+10}{(t+3)(t+4)}=\frac{A}{t+3}+\frac{B}{t+4}$ ⇒ 4t + 10 = A(t + 4) + B(t + 3) ...(i)Putting t = – 3 in (i), we get 4(–3) + 10 = A(–3 + 4) ⇒ A = – 2Putting t = – 4 in (i), we get 4(–4) + 10 = B(–4 + 3) ⇒ B = 6$\therefore\frac{(x^2+1)(x^2+2)}{(x^3+3)(x^2+4)}$ $=1-\left[\frac{-2}{t+3}+\frac{6}{t+4}\right]$ $=1+\left[\frac{2}{t+3}-\frac{6}{t+4}\right]$$\therefore I=\int\left[1+\frac{2}{t+3}-\frac{6}{t+4}\right]dx$$=x+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)-\frac{6}{2}\tan^{-1}\left(\frac{x}{2}\right)+C$$=x+\frac{2}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}-3\tan^{-1}\frac{x}{2}+C$