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

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

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

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

Solution steps

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

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

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

Take the constant out:

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

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

Simplify

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

Expand

x^{2} (3 x + 1) : 3 x^{3} + x^{2}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 3 x^{3} d x = \frac{3 x ^{4}}{4}

\int x^{2} d x = \frac{x ^{3}}{3}

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

Simplify

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

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

Add a constant to the solution

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

y = (x^{3} - 9 x)^{12}, (- 3, 0)

\int (4 sin (x) + 2 cos (x)) d x

x^{2} (1 - 5 x)

\frac{\partial}{\partial u} (a sin (u) cos (v))

f (x) = \frac{216}{x ^{2} + 36}

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