Solve differential equationxu′(x)=u2−4

$x{u}^{\prime }(x)=u^2-4$
$x{u}^{\prime }(x)\frac{1}{u^2-4}=1$
$\frac{1}{u^2-4}du=\frac{1}{x}dx$
Integrating on both sides
$\int \frac{1}{u^2-4}du=\int \frac{1}{x}dx$
The partial fraction of the left side $\frac{1}{(u+2)(u-2)}=-\frac{1}{4(x+2)}+\frac{1}{4(x-2)}$
$-\frac{1}{4}\int \frac{1}{u+2}du+\frac{1}{4}\int \frac{1}{u-2}du=\int \frac{1}{x}dx$
$-\frac{1}{4}\ln (u+2)du+\frac{1}{4}\ln (u-2)=\ln (x)+C_1$
$-\ln (u+2)du+ln(u-2)=4\ln (x)+4C_1$
$\ln (\frac{u-2}{u+2})=\ln (x^4)+C_2$
$(\frac{u-2}{u+2})=(x^4)+e^{C_2}$
$=x^4+C$
$u=-\frac{2(C+x^4)}{-C+x^4}$

Answer is given below (on video)