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

The integral of (3x^2+x/6) is x^3+(x^2)/(12)+C

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

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

x^{3} + \frac{x ^{2}}{12} + C

Solution steps

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

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

\int \frac{x}{6} d x = \frac{x ^{2}}{12}

= x^{3} + \frac{x ^{2}}{12}

Add a constant to the solution

= x^{3} + \frac{x ^{2}}{12} + C

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What is the integral of (3x^2+x/6) ?
The integral of (3x^2+x/6) is x^3+(x^2)/(12)+C