What is the solution for y(x^2+1)^3(dy)/(dx)=x(1-y^2)^2 ?

The solution for y(x^2+1)^3(dy)/(dx)=x(1-y^2)^2 is

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y(x^2+1)^3(dy)/(dx)=x(1-y^2)^2

y (x^{2} + 1)^{3} \frac{d y}{d x} = x (1 - y^{2})^{2}

y = \frac{- ( 4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1 ) ( 2 x ^{4} - 4 c _{1} x ^{4} + 4 x ^{2} - 8 c _{1} x ^{2} + 3 - 4 c _{1} )}{4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1}, y = - \frac{- ( 4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1 ) ( 2 x ^{4} - 4 c _{1} x ^{4} + 4 x ^{2} - 8 c _{1} x ^{2} + 3 - 4 c _{1} )}{4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1}

Solution steps

y (x^{2} + 1)^{3} \frac{d y}{d x} = x (1 - y^{2})^{2}

Solve separable ODE:

y = \frac{- ( 4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1 ) ( 2 x ^{4} - 4 c _{1} x ^{4} + 4 x ^{2} - 8 c _{1} x ^{2} + 3 - 4 c _{1} )}{4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1}, y = - \frac{- ( 4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1 ) ( 2 x ^{4} - 4 c _{1} x ^{4} + 4 x ^{2} - 8 c _{1} x ^{2} + 3 - 4 c _{1} )}{4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1}

y = \frac{- ( 4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1 ) ( 2 x ^{4} - 4 c _{1} x ^{4} + 4 x ^{2} - 8 c _{1} x ^{2} + 3 - 4 c _{1} )}{4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1}, y = - \frac{- ( 4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1 ) ( 2 x ^{4} - 4 c _{1} x ^{4} + 4 x ^{2} - 8 c _{1} x ^{2} + 3 - 4 c _{1} )}{4 c _{1} x ^{4} + 8 c _{1} x ^{2} + 4 c _{1} - 1}

y = \frac{5}{2} - \frac{1}{x}, y = x

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What is the solution for y(x^2+1)^3(dy)/(dx)=x(1-y^2)^2 ?
The solution for y(x^2+1)^3(dy)/(dx)=x(1-y^2)^2 is