What is the integral of x^2sqrt(8+x^3) ?

The integral of x^2sqrt(8+x^3) is 2/9 (8+x^3)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^2sqrt(8+x^3)

\int x^{2} 8 + x^{3} d x

\frac{2}{9} (8 + x^{3})^{\frac{3}{2}} + C

Solution steps

\int x^{2} 8 + x^{3} d x

Apply u-substitution

= \int \frac{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 u d u

Apply the Power Rule

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 8 + x^{3}

= \frac{1}{3} \cdot \frac{2}{3} (8 + x^{3})^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (8 + x^{3})^{\frac{3}{2}} : \frac{2}{9} (8 + x^{3})^{\frac{3}{2}}

= \frac{2}{9} (8 + x^{3})^{\frac{3}{2}}

Add a constant to the solution

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

What is the integral of x^2sqrt(8+x^3) ?
The integral of x^2sqrt(8+x^3) is 2/9 (8+x^3)^{3/2}+C