What is the integral of (6x+2)sqrt(3x^2+2x) ?

The integral of (6x+2)sqrt(3x^2+2x) is 2/(9sqrt(3))(9x^2+6x)^{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 (6x+2)sqrt(3x^2+2x)

\int (6 x + 2) 3 x^{2} + 2 x d x

\frac{2}{9 3} (9 x^{2} + 6 x)^{\frac{3}{2}} + C

Solution steps

\int (6 x + 2) 3 x^{2} + 2 x d x

(6 x + 2) 3 x^{2} + 2 x = 2 (3 x + 1) 3 x^{2} + 2 x

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u u ^{2} - 1}{3 3} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int v^{2} d v

= 2 \cdot \frac{1}{3 3} \cdot \int v^{2} d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

2 \cdot \frac{1}{3 3} \cdot \frac{( ( 3 x + 1 ) ^{2} - 1 ) ^{3}}{3} : \frac{2}{9 3} (9 x^{2} + 6 x)^{\frac{3}{2}}

= \frac{2}{9 3} (9 x^{2} + 6 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9 3} (9 x^{2} + 6 x)^{\frac{3}{2}} + C

\frac{d}{d x} (4 x^{3} - x^{4})

\frac{d y}{d x} = k (10 - y) (5 - y)

\int x^{- 9} d x

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

f (x) = x^{2} + 4 x - 1

What is the integral of (6x+2)sqrt(3x^2+2x) ?
The integral of (6x+2)sqrt(3x^2+2x) is 2/(9sqrt(3))(9x^2+6x)^{3/2}+C