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

The solution for f^'(x)=x^4(x-2)(x+3) is f(x)=(x^7)/7+(x^6)/6-(6x^5)/5+c_{1}

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

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

f (x) = \frac{x ^{7}}{7} + \frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + c_{1}

Solution steps

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

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

then

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

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

\int x^{4} (x - 2) (x + 3) d x = \frac{x ^{7}}{7} + \frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + c_{1}

f (x) = \frac{x ^{7}}{7} + \frac{x ^{6}}{6} - \frac{6 x ^{5}}{5} + c_{1}

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What is the solution for f^'(x)=x^4(x-2)(x+3) ?
The solution for f^'(x)=x^4(x-2)(x+3) is f(x)=(x^7)/7+(x^6)/6-(6x^5)/5+c_{1}