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

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int x^{2} x^{3} + 2 d x

Apply u-substitution

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

= 3 \cdot \int \frac{u}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = x^{3} + 2

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{3} 3 x - 5 d x

\frac{d}{d x} ((x^{2} + 2)^{7})

\frac{d}{d t} (2 arctan (t))

x^{^{'}} = x^{2} e^{- 2 t}

\int_{0}^{1} (x - e^{x})^{2} d x

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