What is the integral of 1+\sqrt[3]{x} ?

The integral of 1+\sqrt[3]{x} is x+3/4 x^{4/3}+C

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integral of 1+\sqrt[3]{x}

\int 1 + 3 x d x

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

Solution steps

\int 1 + 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 1 d x + \int 3 x d x

\int 1 d x = x

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

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

Add a constant to the solution

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

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What is the integral of 1+\sqrt[3]{x} ?
The integral of 1+\sqrt[3]{x} is x+3/4 x^{4/3}+C