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

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

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integral of 1/(4\sqrt[8]{x^2)}

\frac{1}{3} x^{\frac{3}{4}} + C

Solution steps

Take the constant out:

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

assuming

x \geq 0

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{4}{3} x^{\frac{3}{4}}

Simplify

\frac{1}{4} \cdot \frac{4}{3} x^{\frac{3}{4}} : \frac{1}{3} x^{\frac{3}{4}}

= \frac{1}{3} x^{\frac{3}{4}}

Add a constant to the solution

= \frac{1}{3} x^{\frac{3}{4}} + C

\int 4 y^{3} x d y

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f (x) = \frac{x + 1810}{70 ( x + 2 )}

\int x^{2} - 6 x + 13 d x

x^{2} \frac{d y}{d x} - 2 x y = 4 y^{4}

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