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

The solution for f^'(x)=x^3(x^4-1) is f(x)=(x^8)/8-(x^4)/4+c_{1}

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f^'(x)=x^3(x^4-1)

f^{^{'}} (x) = x^{3} (x^{4} - 1)

f (x) = \frac{x ^{8}}{8} - \frac{x ^{4}}{4} + c_{1}

Solution steps

f^{'} (x) = x^{3} (x^{4} - 1)

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

then

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

f (x) = \int x^{3} (x^{4} - 1) d x

\int x^{3} (x^{4} - 1) d x = \frac{x ^{8}}{8} - \frac{x ^{4}}{4} + c_{1}

f (x) = \frac{x ^{8}}{8} - \frac{x ^{4}}{4} + c_{1}

x \to 0 lim (\frac{1}{1 + 5 ^{\frac{2}{x}}})

\int sin (\frac{1}{3} x) d x

\frac{\partial}{\partial x} (2 sin (3 x))

\frac{d}{d x} (cos^{3} (3 x))

f (x) = 3 x + 7

What is the solution for f^'(x)=x^3(x^4-1) ?
The solution for f^'(x)=x^3(x^4-1) is f(x)=(x^8)/8-(x^4)/4+c_{1}