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Solution:

$\therefore \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$

$\Rightarrow 1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)$

$\Rightarrow 1=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+C x+D x^{2}+D$

Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+C=0$

$B+D=0$

$4 A+C=0$

$4 B+D=1$

On solving these equations, we obtain

$A=0, B=\frac{1}{3}, C=0$, and $D=-\frac{1}{3}$

From equation (1), we obtain

$\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{1}{3\left(x^{2}+1\right)}-\frac{1}{3\left(x^{2}+4\right)}$

$\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\frac{1}{3} \int \frac{1}{x^{2}+1} d x-\frac{1}{3} \int \frac{1}{x^{2}+4} d x$

$=\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$

$=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C$