What is the solution for f^'(x)= 1/3 x^4-5x^2+5 ?

The solution for f^'(x)= 1/3 x^4-5x^2+5 is f(x)=(x^5}{15}-\frac{5x^3)/3+5x+c_{1}

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f^'(x)= 1/3 x^4-5x^2+5

f^{^{'}} (x) = \frac{1}{3} x^{4} - 5 x^{2} + 5

f (x) = \frac{x ^{5}}{15} - \frac{5 x ^{3}}{3} + 5 x + c_{1}

Solution steps

f^{'} (x) = \frac{1}{3} x^{4} - 5 x^{2} + 5

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

then

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

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

\int \frac{1}{3} x^{4} - 5 x^{2} + 5 d x = \frac{x ^{5}}{15} - \frac{5 x ^{3}}{3} + 5 x + c_{1}

f (x) = \frac{x ^{5}}{15} - \frac{5 x ^{3}}{3} + 5 x + c_{1}

y = ln (\frac{1}{x})

\frac{d}{d x} (\frac{tan ( x )}{1 - sec ( x )})

x e^{x} - e^{x}

\int_{- \infty}^{0} x e^{- x^{2}} d x

\frac{\partial}{\partial x} (x - 3 y)

What is the solution for f^'(x)= 1/3 x^4-5x^2+5 ?
The solution for f^'(x)= 1/3 x^4-5x^2+5 is f(x)=(x^5}{15}-\frac{5x^3)/3+5x+c_{1}