What is the solution for f^'(x)= 1/((1-x^2)^2) ?

The solution for f^'(x)= 1/((1-x^2)^2) is f(x)= 1/4 ln(x+1)-1/(4(x+1))-1/4 ln(x-1)-1/(4(x-1))+c_{1}

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f^'(x)= 1/((1-x^2)^2)

f^{^{'}} (x) = \frac{1}{( 1 - x ^{2} ) ^{2}}

f (x) = \frac{1}{4} ln (x + 1) - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln (x - 1) - \frac{1}{4 ( x - 1 )} + c_{1}

Solution steps

f^{'} (x) = \frac{1}{( 1 - x ^{2} ) ^{2}}

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

then

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

f (x) = \int \frac{1}{( 1 - x ^{2} ) ^{2}} d x

\int \frac{1}{( 1 - x ^{2} ) ^{2}} d x = \frac{1}{4} ln (x + 1) - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln (x - 1) - \frac{1}{4 ( x - 1 )} + c_{1}

f (x) = \frac{1}{4} ln (x + 1) - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln (x - 1) - \frac{1}{4 ( x - 1 )} + c_{1}

\frac{d}{d x} (2 x^{\frac{1}{4}} (x^{3} - 11))

f (x) = 5 x^{2} + 4

\frac{d}{d x} (\frac{- 2}{2 x + 5})

\frac{\partial}{\partial y} ((x + y) (x^{2} + 2 y^{2}))

\frac{d y}{d t} + e^{t} y = - 24 e^{t}

What is the solution for f^'(x)= 1/((1-x^2)^2) ?
The solution for f^'(x)= 1/((1-x^2)^2) is f(x)= 1/4 ln(x+1)-1/(4(x+1))-1/4 ln(x-1)-1/(4(x-1))+c_{1}