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

The integral of (2x)/(\sqrt[3]{3x^2+6)} is 1/2 (3x^2+6)^{2/3}+C

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integral of (2x)/(\sqrt[3]{3x^2+6)}

\frac{1}{2} (3 x^{2} + 6)^{\frac{2}{3}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= 2 \cdot \frac{1}{6} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 3 x^{2} + 6

= 2 \cdot \frac{1}{6} \cdot \frac{3}{2} (3 x^{2} + 6)^{\frac{2}{3}}

Simplify

2 \cdot \frac{1}{6} \cdot \frac{3}{2} (3 x^{2} + 6)^{\frac{2}{3}} : \frac{1}{2} (3 x^{2} + 6)^{\frac{2}{3}}

= \frac{1}{2} (3 x^{2} + 6)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{2} (3 x^{2} + 6)^{\frac{2}{3}} + C

y = 4 x^{3} - 9 x^{2} - 12 x + 12, (1, - 5)

\frac{d}{d x} (\frac{tan ( x ) - 1}{sec ( x )})

\frac{d}{d x} (2 arccos (\frac{x}{4}))

f (x) = 2 sin (x), at x = \frac{π}{3}

\int x^{2} 6 x + 5 d x

What is the integral of (2x)/(\sqrt[3]{3x^2+6)} ?
The integral of (2x)/(\sqrt[3]{3x^2+6)} is 1/2 (3x^2+6)^{2/3}+C