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

The integral of (3x)/(\sqrt[3]{10-x^2)} is -9/4 (10-x^2)^{2/3}+C

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integral of (3x)/(\sqrt[3]{10-x^2)}

- \frac{9}{4} (10 - 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}}

= 3 (- \frac{1}{2} \cdot \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 10 - x^{2}

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

Simplify

3 (- \frac{1}{2} \cdot \frac{3}{2} (10 - x^{2})^{\frac{2}{3}}) : - \frac{9}{4} (10 - x^{2})^{\frac{2}{3}}

= - \frac{9}{4} (10 - x^{2})^{\frac{2}{3}}

Add a constant to the solution

= - \frac{9}{4} (10 - x^{2})^{\frac{2}{3}} + C

\frac{d}{d x} (\frac{5 x}{4 x ^{2} + 1})

e^{x} sin (x) + cos (x) e^{x}

f (x) = \frac{1}{4 - x ^{2}}

\int_{1}^{4} 3 x - 6 d x

\int_{1}^{2} x (x - 1) d x

What is the integral of (3x)/(\sqrt[3]{10-x^2)} ?
The integral of (3x)/(\sqrt[3]{10-x^2)} is -9/4 (10-x^2)^{2/3}+C