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

The integral of x^2\sqrt[3]{2-5x^3} is -1/20 (2-5x^3)^{4/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2\sqrt[3]{2-5x^3}

\int x^{2} 3 2 - 5 x^{3} d x

- \frac{1}{20} (2 - 5 x^{3})^{\frac{4}{3}} + C

Solution steps

\int x^{2} 3 2 - 5 x^{3} d x

Apply u-substitution

: \int \frac{u 3 2 - 5 u ^{\frac{3}{2}}}{2 x} d u

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

Take the constant out:

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

= \frac{1}{3} \cdot \int 3 2 - 5 u d u

Apply u-substitution

: \int - \frac{3 v}{5} d v

= \frac{1}{3} \cdot \int - \frac{3 v}{5} d v

Take the constant out:

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

= \frac{1}{3} (- \frac{1}{5} \cdot \int 3 v d v)

Apply the Power Rule

: \frac{3}{4} v^{\frac{4}{3}}

= \frac{1}{3} (- \frac{1}{5} \cdot \frac{3}{4} v^{\frac{4}{3}})

Substitute back

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

Simplify

\frac{1}{3} (- \frac{1}{5} \cdot \frac{3}{4} (2 - 5 x^{3})^{\frac{4}{3}}) : - \frac{1}{20} (2 - 5 x^{3})^{\frac{4}{3}}

= - \frac{1}{20} (2 - 5 x^{3})^{\frac{4}{3}}

Add a constant to the solution

= - \frac{1}{20} (2 - 5 x^{3})^{\frac{4}{3}} + C

\int \frac{( x + 2 ) ^{2}}{x ^{3}} d x

\int \frac{9 e ^{9} x}{x} d x

\int x^{13} x^{7} + 1 d x

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

\int (z^{2} - z + 2) d z

What is the integral of x^2\sqrt[3]{2-5x^3} ?
The integral of x^2\sqrt[3]{2-5x^3} is -1/20 (2-5x^3)^{4/3}+C