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}

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

Solution steps

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 = x

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

Add a constant to the solution

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

x \to \frac{3 π}{2} lim (tan (x))

\int x^{3} e^{4 x^{4}} d x

x \to 1 - lim (\frac{1}{x ^{2} + 1})

\frac{\partial}{\partial x} (x^{2} e^{2 y})

\int x^{3} (x^{2} - 2 x) d x

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