What is the integral of (x^3+\sqrt[3]{x}) ?

The integral of (x^3+\sqrt[3]{x}) is (x^4)/4+3/4 x^{4/3}+C

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integral of (x^3+\sqrt[3]{x})

\int (x^{3} + 3 x) d x

\frac{x ^{4}}{4} + \frac{3}{4} x^{\frac{4}{3}} + C

Solution steps

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

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

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

= \frac{x ^{4}}{4} + \frac{3}{4} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{3}{4} x^{\frac{4}{3}} + C

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What is the integral of (x^3+\sqrt[3]{x}) ?
The integral of (x^3+\sqrt[3]{x}) is (x^4)/4+3/4 x^{4/3}+C