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

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

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

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

\frac{x ^{3}}{6} + \frac{3 x ^{2}}{2} + C

Solution steps

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

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

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

= \frac{x ^{3}}{6} + \frac{3 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{3}}{6} + \frac{3 x ^{2}}{2} + C

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