What is the integral of 5x\sqrt[3]{(9-4x^2)^2} ?

The integral of 5x\sqrt[3]{(9-4x^2)^2} is -3/8 (9-4x^2)^{5/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5x\sqrt[3]{(9-4x^2)^2}

- \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{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

: \int - \frac{u ^{\frac{2}{3}}}{8} d u

= 5 \cdot \int - \frac{u ^{\frac{2}{3}}}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 (- \frac{1}{8} \cdot \int u^{\frac{2}{3}} d u)

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= 5 (- \frac{1}{8} \cdot \frac{3}{5} u^{\frac{5}{3}})

Substitute back

u = 9 - 4 x^{2}

= 5 (- \frac{1}{8} \cdot \frac{3}{5} (9 - 4 x^{2})^{\frac{5}{3}})

Simplify

5 (- \frac{1}{8} \cdot \frac{3}{5} (9 - 4 x^{2})^{\frac{5}{3}}) : - \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}}

= - \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}}

Add a constant to the solution

= - \frac{3}{8} (9 - 4 x^{2})^{\frac{5}{3}} + C

\int sec (x) tan (x) 1 + sec (x) d x

f (x) = x ∣ x ∣

\int_{4}^{- 2} \frac{1}{2} d x

x^{^{'}} = x + 3

9 x - 6 x

What is the integral of 5x\sqrt[3]{(9-4x^2)^2} ?
The integral of 5x\sqrt[3]{(9-4x^2)^2} is -3/8 (9-4x^2)^{5/3}+C