What is the solution for f^'(x)= 4/(1-2x^2) ?

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

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f^'(x)= 4/(1-2x^2)

f^{^{'}} (x) = \frac{4}{1 - 2 x ^{2}}

f (x) = 2 ln (2 x + 1) - 2 ln (2 x - 1) + c_{1}

Solution steps

f^{'} (x) = \frac{4}{1 - 2 x ^{2}}

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

then

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

f (x) = \int \frac{4}{1 - 2 x ^{2}} d x

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

f (x) = 2 ln (2 x + 1) - 2 ln (2 x - 1) + c_{1}

sin (t^{2}) ln (t^{3})

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

\int 4 e^{x} d x

x + y^{2} = 30, x + y = 0

\int \frac{y ^{5}}{( 1 - y ^{3} ) ^{\frac{3}{2}}} d y

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