$\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
$\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
Let $\frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+3\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$
$4 x^{2}+10=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+3\right)$
$4 x^{2}+10=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+3 C x+D x^{2}+3 D$
$4 x^{2}+10=(A+C) x^{3}+(B+D) x^{2}+(4 A+3 C) x+(4 B+3 D)$
Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain
$A+C=0$
$B+D=4$
$4 A+3 C=0$
$4 B+3 D=10$
On solving these equations, we obtain
A = 0, B = −2, C = 0, and D = 6
$\therefore \frac{4 x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}$
$\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\left(\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}\right)$
$\Rightarrow \int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x=\int\left\{1+\frac{2}{\left(x^{2}+3\right)}-\frac{6}{\left(x^{2}+4\right)}\right\} d x$
$=\int\left\{1+\frac{2}{x^{2}+(\sqrt{3})^{2}}-\frac{6}{x^{2}+2^{2}}\right\}$
$=x+2\left(\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2} \tan ^{-1} \frac{x}{2}\right)+\mathrm{C}$
$=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+\mathrm{C}$