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

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (ln (3 x + 4 y))

\frac{\partial}{\partial x} (x^{3} - 9 x y + y^{3} - 2)

\int_{1}^{e} \frac{sec ( 2 + ln ( x ) )}{x} d x

\int e^{2 x + l n (x)} d x

x \to 7 - lim (\frac{1}{x ^{2} ( x + 7 )})

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