What is the integral of x^2(sqrt(4+x^3)) ?

The integral of x^2(sqrt(4+x^3)) is 2/9 (4+x^3)^{3/2}+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(4+x^3))

\int x^{2} (4 + x^{3}) d x

\frac{2}{9} (4 + x^{3})^{\frac{3}{2}} + C

Solution steps

\int x^{2} 4 + x^{3} d x

Apply u-substitution

: \int \frac{u}{3} d u

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

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

Substitute back

u = 4 + x^{3}

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

Simplify

\frac{1}{3} \cdot \frac{2}{3} (4 + x^{3})^{\frac{3}{2}} : \frac{2}{9} (4 + x^{3})^{\frac{3}{2}}

= \frac{2}{9} (4 + x^{3})^{\frac{3}{2}}

Add a constant to the solution

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

x y^{^{'}} + y = 5 x y^{2}

\int \frac{1}{( 6 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (2 x - \frac{2000}{x ^{2}})

\int 6 arctan (4 y) d y

\frac{\partial}{\partial x} (2 y^{4 x})

What is the integral of x^2(sqrt(4+x^3)) ?
The integral of x^2(sqrt(4+x^3)) is 2/9 (4+x^3)^{3/2}+C