What is the integral of (x^3)/3-(3x^2)/2+23/6 ?

The integral of (x^3)/3-(3x^2)/2+23/6 is (x^4}{12}-\frac{x^3)/2+23/6 x+C

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integral of (x^3)/3-(3x^2)/2+23/6

\int \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + \frac{23}{6} d x

\frac{x ^{4}}{12} - \frac{x ^{3}}{2} + \frac{23}{6} x + C

Solution steps

\int \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + \frac{23}{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 \frac{x ^{3}}{3} d x - \int \frac{3 x ^{2}}{2} d x + \int \frac{23}{6} d x

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

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

\int \frac{23}{6} d x = \frac{23}{6} x

= \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + \frac{23}{6} x

Add a constant to the solution

= \frac{x ^{4}}{12} - \frac{x ^{3}}{2} + \frac{23}{6} x + C

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What is the integral of (x^3)/3-(3x^2)/2+23/6 ?
The integral of (x^3)/3-(3x^2)/2+23/6 is (x^4}{12}-\frac{x^3)/2+23/6 x+C