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

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

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

f^{^{'}} (x) = x^{2} - x - 2

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

Solution steps

f^{'} (x) = x^{2} - x - 2

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

then

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

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

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

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

d er i v a t i v e 6 e^{x}

\frac{d y}{d x} = 5 y + a

\int (x^{2} + 4 x) d x

in v erse l a pl a ce \frac{4}{( s + 1 ) ( s + 2 ) ^{3}}

\frac{d}{d x} (\frac{5 x + 2}{3 x - 4})

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