What is the solution for f^'(x)=x^{1/3}(2x-1)(x+1) ?

The solution for f^'(x)=x^{1/3}(2x-1)(x+1) is f(x)= 3/5 x^{10/3}+3/7 x^{7/3}-3/4 x^{4/3}+c_{1}

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f^'(x)=x^{1/3}(2x-1)(x+1)

f^{^{'}} (x) = x^{\frac{1}{3}} (2 x - 1) (x + 1)

f (x) = \frac{3}{5} x^{\frac{10}{3}} + \frac{3}{7} x^{\frac{7}{3}} - \frac{3}{4} x^{\frac{4}{3}} + c_{1}

Solution steps

f^{'} (x) = x^{\frac{1}{3}} (2 x - 1) (x + 1)

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

then

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

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

\int x^{\frac{1}{3}} (2 x - 1) (x + 1) d x = \frac{3}{5} x^{\frac{10}{3}} + \frac{3}{7} x^{\frac{7}{3}} - \frac{3}{4} x^{\frac{4}{3}} + c_{1}

f (x) = \frac{3}{5} x^{\frac{10}{3}} + \frac{3}{7} x^{\frac{7}{3}} - \frac{3}{4} x^{\frac{4}{3}} + c_{1}

f (x) = 2 x + 1, at x = \frac{1}{4}

\int 4 e^{\frac{x}{2}} d x

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

\frac{d}{d x} (\frac{x ^{e^{x}}}{1 + x ^{2}})

\frac{d}{d x} (sin (e^{x}))

What is the solution for f^'(x)=x^{1/3}(2x-1)(x+1) ?
The solution for f^'(x)=x^{1/3}(2x-1)(x+1) is f(x)= 3/5 x^{10/3}+3/7 x^{7/3}-3/4 x^{4/3}+c_{1}