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

The integral of ((\sqrt[3]{t})^2-2) is 3/5 t^{5/3}-2t+C

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integral of ((\sqrt[3]{t})^2-2)

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

\frac{3}{5} t^{\frac{5}{3}} - 2 t + C

Solution steps

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

Apply the Sum Rule:

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

= \int (3 t)^{2} d t - \int 2 d t

\int (3 t)^{2} d t = \frac{3}{5} t^{\frac{5}{3}}

\int 2 d t = 2 t

= \frac{3}{5} t^{\frac{5}{3}} - 2 t

Add a constant to the solution

= \frac{3}{5} t^{\frac{5}{3}} - 2 t + C

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What is the integral of ((\sqrt[3]{t})^2-2) ?
The integral of ((\sqrt[3]{t})^2-2) is 3/5 t^{5/3}-2t+C