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

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

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(dy)/(dx)= 1/(x-x^3)

\frac{d y}{d x} = \frac{1}{x - x ^{3}}

y = ln (x) - \frac{1}{2} ln (x + 1) - \frac{1}{2} ln (x - 1) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{1}{x - x ^{3}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{1}{x - x ^{3}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1}{x - x ^{3}} d x

\int \frac{1}{x - x ^{3}} d x = ln (x) - \frac{1}{2} ln (x + 1) - \frac{1}{2} ln (x - 1) + c_{1}

y = ln (x) - \frac{1}{2} ln (x + 1) - \frac{1}{2} ln (x - 1) + c_{1}

\frac{\partial}{\partial x} (2 x - 3 y + 5 z)

\int 3^{2 x} cos^{2} (3^{x}) d x

f (x) = x - 3

2 x \frac{d y}{d x} + y = 10 x

\int e^{- x} sec^{2} (e^{- x}) d x

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