What is the integral of (x-1)/(\sqrt[3]{x^2)} ?

The integral of (x-1)/(\sqrt[3]{x^2)} is 3/4 x^{4/3}-3\sqrt[3]{x}+C

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integral of (x-1)/(\sqrt[3]{x^2)}

Solution steps

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

Add a constant to the solution

\int (x + 1) (x^{2} + 2 x)^{3} d x

\int \frac{x ^{5} - 5 x}{x ^{4}} d x

- sin (x)

\int sec^{2} (10 x) d x

\int 2 e^{x} + 3 d x

What is the integral of (x-1)/(\sqrt[3]{x^2)} ?
The integral of (x-1)/(\sqrt[3]{x^2)} is 3/4 x^{4/3}-3\sqrt[3]{x}+C