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

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

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

\int \frac{1}{3 t ^{2}} d t

3 3 t + C

Solution steps

\int \frac{1}{3 t ^{2}} d t

3 t^{2} = t^{\frac{2}{3}},

assuming

t \geq 0

= \int \frac{1}{t ^{\frac{2}{3}}} d t

Apply the Power Rule

: 3 t^{\frac{1}{3}}

= 3 t^{\frac{1}{3}}

Apply radical rule:

a^{\frac{1}{n}} = n a

= 3 3 t

Add a constant to the solution

= 3 3 t + C

\int \frac{e ^{a} + 1}{e ^{a}} d a

x \to - 8 lim (\frac{x ^{2} - 4}{8 - x})

\int \frac{1}{4 e ^{- 4 x} + e ^{4 x}} d x

x \to 0 lim (f (x) (x^{2}))

\frac{d x}{d t} = 1

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