Solve the following differential equation: (x²-1)dy/dx + 2xy = 2x²

(x²-1)dy/dx + 2xy = 2x²
dy/dx + 2x/(x²-1)y = 2x²/(x²-1)
It is in the form of linear differential equation.
dy/dx + Py = Q
Where P = 2x/(x²-1), Q = 2x²/(x²-1)
Integrating factor, IF = e∫Pdx = e∫2x/(x²-1)dx = e^(ln|x²-1|) = |x²-1|
Therefore, y|x²-1| = ∫2x²|x²-1|/ (x²-1) dx + C
Assuming x² > 1, y(x²-1) = ∫2x² dx + C = (2/3)x³ + C
y = [(2/3)x³ + C]/(x²-1)
Assuming x² < 1, y(1-x²) = ∫-2x² dx + C = (-2/3)x³ + C
y = [(-2/3)x³ + C]/(1-x²)