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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(\sqrt[3]{9-x^2)}

- \frac{3}{2} (9 - 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}{2} \cdot \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 9 - x^{2}

= 2 (- \frac{1}{2} \cdot \frac{3}{2} (9 - x^{2})^{\frac{2}{3}})

Simplify

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

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

Add a constant to the solution

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

x \to 1 - lim (x^{(\frac{1}{1 - x})})

\frac{d}{d z} (x y e^{z})

f (x) = ln (x + 9)

\frac{d y}{d x} = e^{3 x} + 2 y

\int_{1}^{10} \frac{13}{x ^{3}} d x

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