1st order differential equation help

du/dx = (u^2-4)/x - does anyone know how to get the general solution from that? I'd like to see full method.

If that helps, 4/(u^2-4) can be written as 1/(u-2)-1/(u+2) - that was the first part of the question.

Find the general solution by separating the variables then integrating:
du / dx = (u² - 4) / x
du / (u² - 4) = dx / x
[1 / {4(u - 2)} - 1 / {4(u + 2)}] du = dx / x
∫ [1 / {4(u - 2)} - 1 / {4(u + 2)}] du = ∫ 1 / x dx
ln|u - 2| / 4 - ln|u + 2| / 4 = ln|x| + C
ln|u - 2| / 4 - ln|u + 2| / 4 = C + ln|x|
ln|u - 2| - ln|u + 2| = C + 4ln|x|
ln|(u - 2) / (u + 2)| = C + 4ln|x|
(u - 2) / (u + 2) = ℮^(C + 4ln|x|)
(u - 2) / (u + 2) = ℮^(C + lnx⁴)
(u - 2) / (u + 2) = ℮ᶜx⁴
(u - 2) / (u + 2) = Cx⁴
u - 2 = Cx⁴(u + 2)
u - 2 = Cux⁴ + 2Cx⁴
u - Cux⁴ = 2Cx⁴ + 2
u(1 - Cx⁴) = 2Cx⁴ + 2
u = (2Cx⁴ + 2) / (1 - Cx⁴)
u = 2(Cx⁴ + 1) / (1 - Cx⁴)

du/(u^2 - 4) = dx/x

Integrating both sides:

1/4*[ln|(u - 2)/(u + 2)|] = ln|x| + C

solve for u.