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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\frac{3}{2} (x^{3} - 2)^{\frac{4}{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 u-substitution

Apply the Power Rule

= 6 \cdot \frac{1}{3} \cdot \frac{3}{4} v^{\frac{4}{3}}

Substitute back

= 6 \cdot \frac{1}{3} \cdot \frac{3}{4} (x^{3} - 2)^{\frac{4}{3}}

Simplify

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

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

Add a constant to the solution

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

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