What is the solution for f^'(x)=x^4+3x^3+5x^2+2x ?

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

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f^'(x)=x^4+3x^3+5x^2+2x

f^{^{'}} (x) = x^{4} + 3 x^{3} + 5 x^{2} + 2 x

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

Solution steps

f^{'} (x) = x^{4} + 3 x^{3} + 5 x^{2} + 2 x

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

then

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

f (x) = \int x^{4} + 3 x^{3} + 5 x^{2} + 2 x d x

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

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

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