$x{u}^{\prime}(x)={u}^{2}-4$

$x{u}^{\prime}(x)\frac{1}{{u}^{2}-4}=1$

$\frac{1}{{u}^{2}-4}du=\frac{1}{x}dx$

Integrating on both sides

$\int \frac{1}{{u}^{2}-4}du=\int \frac{1}{x}dx$

The partial fraction of the left side $\frac{1}{(u+2)(u-2)}=-\frac{1}{4(x+2)}+\frac{1}{4(x-2)}$

$-\frac{1}{4}\int \frac{1}{u+2}du+\frac{1}{4}\int \frac{1}{u-2}du=\int \frac{1}{x}dx$

$-\frac{1}{4}\mathrm{ln}(u+2)du+\frac{1}{4}\mathrm{ln}(u-2)=\mathrm{ln}(x)+{C}_{1}$

$-\mathrm{ln}(u+2)du+ln(u-2)=4\mathrm{ln}(x)+4{C}_{1}$

$\mathrm{ln}(\frac{u-2}{u+2})=\mathrm{ln}({x}^{4})+{C}_{2}$

$(\frac{u-2}{u+2})=({x}^{4})+{e}^{{C}_{2}}$

$={x}^{4}+C$

$u=-\frac{2(C+{x}^{4})}{-C+{x}^{4}}$