What is the integral of sqrt(\sqrt[6]{x^4)} ?

The integral of sqrt(\sqrt[6]{x^4)} is 3/4 x^{4/3}+C

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integral of sqrt(\sqrt[6]{x^4)}

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

Solution steps

Apply u-substitution

: \int \frac{1}{4 u ^{\frac{2}{3}}} d u

= \int \frac{1}{4 u ^{\frac{2}{3}}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

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

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

Substitute back

u = x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d v}{d t} + \frac{( v ^{^{'}} \cdot v )}{t} = v^{^{'}} \cdot (\frac{t}{2} - \frac{1}{2})

\frac{d}{d x} (2 x ln (x))

x \to 0 lim (3 x^{2} + 2 x h + h^{2})

\frac{\partial}{\partial x} (9 - x^{2} - y)

\frac{\partial}{\partial x} (- (1 + 3 x y))

What is the integral of sqrt(\sqrt[6]{x^4)} ?
The integral of sqrt(\sqrt[6]{x^4)} is 3/4 x^{4/3}+C