What is the solution for f^'(x)=x^4-3x^3-6x^2+6x-1 ?

The solution for f^'(x)=x^4-3x^3-6x^2+6x-1 is f(x)=(x^5)/5-(3x^4)/4-2x^3+3x^2-x+c_{1}

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f^'(x)=x^4-3x^3-6x^2+6x-1

f^{^{'}} (x) = x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1

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

Solution steps

f^{'} (x) = x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1

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

then

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

f (x) = \int x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1 d x

\int x^{4} - 3 x^{3} - 6 x^{2} + 6 x - 1 d x = \frac{x ^{5}}{5} - \frac{3 x ^{4}}{4} - 2 x^{3} + 3 x^{2} - x + c_{1}

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

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What is the solution for f^'(x)=x^4-3x^3-6x^2+6x-1 ?
The solution for f^'(x)=x^4-3x^3-6x^2+6x-1 is f(x)=(x^5)/5-(3x^4)/4-2x^3+3x^2-x+c_{1}