What is the integral of 1/(\sqrt[5]{t^4)} ?

The integral of 1/(\sqrt[5]{t^4)} is 5\sqrt[5]{t}+C

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integral of 1/(\sqrt[5]{t^4)}

\int \frac{1}{5 t ^{4}} d t

5 5 t + C

Solution steps

\int \frac{1}{5 t ^{4}} d t

5 t^{4} = t^{\frac{4}{5}},

assuming

t \geq 0

= \int \frac{1}{t ^{\frac{4}{5}}} d t

Apply the Power Rule

: 5 t^{\frac{1}{5}}

= 5 t^{\frac{1}{5}}

Apply radical rule:

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

= 5 5 t

Add a constant to the solution

= 5 5 t + C

x sin (x^{2})

\int \frac{4}{x 1 + 9 x ^{2}} d x

\int \frac{1}{14 - x ^{2}} d x

\int t^{- 5} d t

\int \frac{in x}{2 x} d x

What is the integral of 1/(\sqrt[5]{t^4)} ?
The integral of 1/(\sqrt[5]{t^4)} is 5\sqrt[5]{t}+C