What is the integral of-(x^3+2)/6 ?

The integral of-(x^3+2)/6 is -1/6 ((x^4)/4+2x)+C

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integral of-(x^3+2)/6

\int - \frac{x ^{3} + 2}{6} d x

- \frac{1}{6} (\frac{x ^{4}}{4} + 2 x) + C

Solution steps

\int - \frac{x ^{3} + 2}{6} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{6} \cdot \int x^{3} + 2 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

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

\int x^{3} d x = \frac{x ^{4}}{4}

\int 2 d x = 2 x

= - \frac{1}{6} (\frac{x ^{4}}{4} + 2 x)

Add a constant to the solution

= - \frac{1}{6} (\frac{x ^{4}}{4} + 2 x) + C

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What is the integral of-(x^3+2)/6 ?
The integral of-(x^3+2)/6 is -1/6 ((x^4)/4+2x)+C