What is the integral of ((x^2)/3-8x-3) ?

The integral of ((x^2)/3-8x-3) is (x^3)/9-4x^2-3x+C

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integral of ((x^2)/3-8x-3)

\int (\frac{x ^{2}}{3} - 8 x - 3) d x

\frac{x ^{3}}{9} - 4 x^{2} - 3 x + C

Solution steps

\int \frac{x ^{2}}{3} - 8 x - 3 d x

Apply the Sum Rule:

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

= \int \frac{x ^{2}}{3} d x - \int 8 x d x - \int 3 d x

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

\int 8 x d x = 4 x^{2}

\int 3 d x = 3 x

= \frac{x ^{3}}{9} - 4 x^{2} - 3 x

Add a constant to the solution

= \frac{x ^{3}}{9} - 4 x^{2} - 3 x + C

What is the integral of ((x^2)/3-8x-3) ?
The integral of ((x^2)/3-8x-3) is (x^3)/9-4x^2-3x+C