Let, $I=\int \frac{5 x^2+3}{x^2\left(x^2-2\right)} d x$
Put $x^2=y$. Then,


$$
\begin{aligned}
& 5 y+3=(y-2) A+y B \\
& 5 y+3=y(A+B)-2 A
\end{aligned}
$$
On comparing the coefficients, we get $A+B=5$ and $3=-2 A$
$A=\frac{-3}{2}$ and $B=\frac{13}{2}$
Thus, $\frac{5 y+3}{y(y-2)}=\frac{5 x^2+3}{x^2\left(x^2-2\right)}=-\frac{3}{2} \frac{1}{x^2}+\frac{13}{2} \frac{1}{x^2-2}$
Now, $I=-\frac{3}{2} \int \frac{d x}{x^2}+\frac{13}{2} \int \frac{d x}{x^2-2}$
$$
\begin{gathered}
=-\frac{3}{2}\left(-\frac{1}{x}\right)+\frac{13}{2} \times \frac{1}{2 \sqrt{2}} \log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C \\
=\frac{3}{2 x}+\frac{\sqrt{13}}{4 \sqrt{2}} \log \left|\frac{x-\sqrt{2}}{x+\sqrt{2}}\right|+C
\end{gathered}
$$