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

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

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f^'(x)=2x^2+3x-2

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

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

Solution steps

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

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

then

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

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

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

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

\frac{\partial}{\partial x} (x^{4} y - x^{2} y^{3})

y = 3 x^{2} + \frac{4 x}{3} - 2

\frac{3}{r ^{3}}

\frac{d y}{d x} = \frac{1 - y ^{2}}{x}

n = 6 \sum \infty \frac{5 ^{n}}{1 0 ^{n}}

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