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

The integral of ((2x^4)/(\sqrt[3]{x^2)}) is 6/13 x^{13/3}+C

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integral of ((2x^4)/(\sqrt[3]{x^2)})

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

\frac{6}{13} x^{\frac{13}{3}} + C

Solution steps

\int \frac{2 x ^{4}}{3 x ^{2}} d x

Take the constant out:

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

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

3 x^{2} = x^{\frac{2}{3}},

assuming

x \geq 0

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

Simplify

\frac{x ^{4}}{x ^{\frac{2}{3}}} : x^{\frac{10}{3}}

= 2 \cdot \int x^{\frac{10}{3}} d x

Apply the Power Rule

: \frac{3}{13} x^{\frac{13}{3}}

= 2 \cdot \frac{3}{13} x^{\frac{13}{3}}

Simplify

2 \cdot \frac{3}{13} x^{\frac{13}{3}} : \frac{6}{13} x^{\frac{13}{3}}

= \frac{6}{13} x^{\frac{13}{3}}

Add a constant to the solution

= \frac{6}{13} x^{\frac{13}{3}} + C

y^{^{''}} + 36 y = - 36 sec^{2} (6 x)

f (x) = x + 6

x = 7 y, \frac{1}{3} y = x, \frac{1}{7} y = x

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

x \to - 7 lim (\frac{x ^{2} - 8}{7 - x})

What is the integral of ((2x^4)/(\sqrt[3]{x^2)}) ?
The integral of ((2x^4)/(\sqrt[3]{x^2)}) is 6/13 x^{13/3}+C