Example 13. Find $\int \frac{7+x^{2}}{x^{2}\left(x^{2}+2\right)} d x$.
Solution. The fraction can be decomposed into two fractions:

$$
\frac{7}{x^{2}\left(x^{2}+2\right)}+\frac{x^{2}}{x^{2}\left(x^{2}+2\right)}=\frac{7}{x^{2}\left(x^{2}+2\right)}+\frac{1}{x^{2}+2}
$$

What to do with the right fraction, which does not fit the tabulated formulas? We can obtain a fraction with the denominator \(x^{2}\left(x^{2}+2\right)\) by adding or subtracting two fractions: \(\frac{1}{x^{2}}\) and \(\frac{1}{x^{2}+2}\). When adding, we get \(\frac{2 x^{2}+2}{x^{2}\left(x^{2}+2\right)}\), which does not lead to the required fraction. Subtraction: \(\frac{1}{x^{2}}-\frac{1}{x^{2}+2}=\frac{2}{x^{2}\left(x^{2}+2\right)}\). This attempt is successful, as \(\frac{7}{x^{2}\left(x^{2}+2\right)}=\frac{7}{2}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}+2}\right)\) (here is the trick: the fraction is decomposed into simpler fractions that are listed in the table). Let's implement this idea.

$$
\begin{aligned}
& \int \frac{7+x^{2}}{x^{2}\left(x^{2}+2\right)} d x= \\
& =\left\{\begin{array}{r}
\frac{7}{x^{2}(x+2)}+\frac{1}{x^{2}+2}=\frac{7}{2} \cdot \frac{1}{x^{2}}-\frac{7}{2} \cdot \frac{1}{x^{2}+2}+\frac{1}{x^{2}+2}= \\
=\frac{7}{2} \cdot \frac{1}{x^{2}}-\frac{5}{2} \cdot \frac{1}{x^{2}+2}
\end{array}\right\}= \\
& =\frac{7}{2} \int \frac{d x}{x^{2}}-\frac{5}{2} \int \frac{d x}{x^{2}+(\sqrt{2})^{2}}=-\frac{7}{2 x}-\frac{5}{2 \sqrt{2}} \operatorname{arctg} \frac{x}{\sqrt{2}}+C
\end{aligned}
$$