\(xu'(x)= u^2-4\)

\(xu'(x) \frac{1}{u^2-4}=1\)

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

Integrating on both sides

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

The partial fraction of the left side \(\frac{1}{(u+2)(uā2)}= ā\frac{1}{4(x+2)}+\frac{1}{4(xā2)}\)

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

\(-\frac{1}{4} \ln (u+2)du+\frac{1}{4} \ln(u-2)= \ln(x)+C_1\)

\(-\ln(u+2)du+ln(u-2)= 4 \ln(x)+4C_1\)

\(\ln(\frac{u-2}{u+2})= \ln(x^4)+C_2\)

\((\frac{u-2}{u+2})= (x^4)+e^{C_2}\)

\(= x^4+C\)

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