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

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

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integral of (x^3)/3+x

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

\frac{x ^{4}}{12} + \frac{x ^{2}}{2} + C

Solution steps

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

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

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

= \frac{x ^{4}}{12} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{12} + \frac{x ^{2}}{2} + C

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