I think you mean x(1-x2)y'+(2x2-1)y=x3.
Divide through by x(1-x2):
y'+(2x2-1)y/(x(1-x2))=x2/(1-x2).
We can write (2x2-1)/(x(1-x2))=A/x+B/(1-x)+C/(1+x).
So 2x2-1=A-Ax2+Bx+Bx2+Cx-Cx2. Equating coefficients:
Constant term: -1=A, A=-1;
x term: 0=B+C, so C=-B;
x2 term: 2=-A+B-C=1+2B, B=½, C=-½.
(2x2-1)/(x(1-x2))=-1/x+½/(1-x)-½/(1+x).
Integrate this wrt x: ln(1/|x|)-½ln|1-x|-½ln|1+x|=ln|1/(x√(1-x2))|, now we can work out an integrating factor:
eln|1/(x√(1-x²))|=1/(x√(1-x2)).
Multiply through by this factor:
(1/(x√(1-x2)))y'+(1/(x√(1-x2)))((2x2-1)y/(x(1-x2)))=(1/(x√(1-x2)))(x2/(1-x2)).
(d/dx)(y/(x√(1-x2)))=x/(1-x2)3/2.
y/(x√(1-x2))=∫xdx/(1-x2)3/2.
Let u=1-x2, then du=-2xdx, so xdx=-½du. The integrand becomes: u-3/2(-½du).
-½∫u-3/2du=1/√u=1/√(1-x2)+C, where C is integration constant.
y/(x√(1-x2))=1/√(1-x2)+C, y=(x√(1-x2))/√(1-x2)+Cx√(1-x2)=x+Cx√(1-x2).