What is the integral of 1/(\sqrt[8]{x^2)} ?

The integral of 1/(\sqrt[8]{x^2)} is 4/3 x^{3/4}+C

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integral of 1/(\sqrt[8]{x^2)}

\frac{4}{3} x^{\frac{3}{4}} + C

Solution steps

assuming

x \geq 0

Apply the Power Rule

: \frac{4}{3} x^{\frac{3}{4}}

= \frac{4}{3} x^{\frac{3}{4}}

Add a constant to the solution

= \frac{4}{3} x^{\frac{3}{4}} + C

y^{^{'}} = - \frac{π}{2} t (1 + y^{2})

\frac{d}{d x} (\frac{ln ( x + 1 )}{x + 1})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} + 1}{x ^{2} - 1})

\frac{\partial}{\partial x} ((x - 4)^{2} + y^{2})

(2 x + y cos (x y)) d x + x cos (x y) d y = 0

What is the integral of 1/(\sqrt[8]{x^2)} ?
The integral of 1/(\sqrt[8]{x^2)} is 4/3 x^{3/4}+C