What is the integral of 2x\sqrt[4]{1-x^2} ?

The integral of 2x\sqrt[4]{1-x^2} is -4/5 (1-x^2)^{5/4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2x\sqrt[4]{1-x^2}

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

= 2 (- \frac{1}{2} \cdot \frac{4}{5} u^{\frac{5}{4}})

Substitute back

u = 1 - x^{2}

= 2 (- \frac{1}{2} \cdot \frac{4}{5} (1 - x^{2})^{\frac{5}{4}})

Simplify

2 (- \frac{1}{2} \cdot \frac{4}{5} (1 - x^{2})^{\frac{5}{4}}) : - \frac{4}{5} (1 - x^{2})^{\frac{5}{4}}

= - \frac{4}{5} (1 - x^{2})^{\frac{5}{4}}

Add a constant to the solution

= - \frac{4}{5} (1 - x^{2})^{\frac{5}{4}} + C

What is the integral of 2x\sqrt[4]{1-x^2} ?
The integral of 2x\sqrt[4]{1-x^2} is -4/5 (1-x^2)^{5/4}+C