What is the integral of (2x)/(\sqrt[3]{6-5x^2)} ?

The integral of (2x)/(\sqrt[3]{6-5x^2)} is -3/10 (6-5x^2)^{2/3}+C

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integral of (2x)/(\sqrt[3]{6-5x^2)}

- \frac{3}{10} (6 - 5 x^{2})^{\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 (- \frac{1}{10} \cdot \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 6 - 5 x^{2}

= 2 (- \frac{1}{10} \cdot \frac{3}{2} (6 - 5 x^{2})^{\frac{2}{3}})

Simplify

2 (- \frac{1}{10} \cdot \frac{3}{2} (6 - 5 x^{2})^{\frac{2}{3}}) : - \frac{3}{10} (6 - 5 x^{2})^{\frac{2}{3}}

= - \frac{3}{10} (6 - 5 x^{2})^{\frac{2}{3}}

Add a constant to the solution

= - \frac{3}{10} (6 - 5 x^{2})^{\frac{2}{3}} + C

f (x) =, 4 x^{3} + 2 t + 9

f (x) = (2 x + 3)^{1.4}

\int \frac{( x ^{2} + 1 ) ( x ^{2} - 2 )}{x ^{\frac{2}{3}}} d x

y = x + 2, x = - 2, y = 2

f (x) = 11 x^{2} + 6 x, at x = 2

What is the integral of (2x)/(\sqrt[3]{6-5x^2)} ?
The integral of (2x)/(\sqrt[3]{6-5x^2)} is -3/10 (6-5x^2)^{2/3}+C