What is the integral of 8/(\sqrt[3]{x^4)} ?

The integral of 8/(\sqrt[3]{x^4)} is -(24)/(\sqrt[3]{x)}+C

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integral of 8/(\sqrt[3]{x^4)}

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

assuming

x \geq 0

= 8 \cdot \int \frac{1}{x ^{\frac{4}{3}}} d x

Apply the Power Rule

Simplify

Add a constant to the solution

\int \frac{1}{x ^{α}} d x

\int \frac{sin ( ln ( t ) )}{t} d t

\int \frac{3}{4} x^{\frac{4}{3}} d x

\int 56 x^{- \frac{1}{5}} tan (x^{\frac{4}{5}}) d x

\int (4 x^{3} + 3) (2 x^{4} + 6 x)^{\frac{1}{3}} d x

What is the integral of 8/(\sqrt[3]{x^4)} ?
The integral of 8/(\sqrt[3]{x^4)} is -(24)/(\sqrt[3]{x)}+C