What is the integral of 12*(2x+7)*(x+1)^2 ?

The integral of 12*(2x+7)*(x+1)^2 is 12((x^4)/2+(11x^3)/3+8x^2+7x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 12(2x+7)(x+1)^2

\int 12 \cdot (2 x + 7) \cdot (x + 1)^{2} d x

12 (\frac{x ^{4}}{2} + \frac{11 x ^{3}}{3} + 8 x^{2} + 7 x) + C

Solution steps

\int 12 (2 x + 7) (x + 1)^{2} d x

Take the constant out:

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

= 12 \cdot \int (2 x + 7) (x + 1)^{2} d x

Expand

(2 x + 7) (x + 1)^{2} : 2 x^{3} + 11 x^{2} + 16 x + 7

= 12 \cdot \int 2 x^{3} + 11 x^{2} + 16 x + 7 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

= 12 (\int 2 x^{3} d x + \int 11 x^{2} d x + \int 16 x d x + \int 7 d x)

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 11 x^{2} d x = \frac{11 x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 7 d x = 7 x

= 12 (\frac{x ^{4}}{2} + \frac{11 x ^{3}}{3} + 8 x^{2} + 7 x)

Add a constant to the solution

= 12 (\frac{x ^{4}}{2} + \frac{11 x ^{3}}{3} + 8 x^{2} + 7 x) + C

x \to 0 + lim (x^{4} ln (x))

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

6 cos (4 π x)

\frac{d y}{d x} + 3 x^{2} y = 9 x^{2}

\frac{d}{d x} (7 sin^{3} (x))

What is the integral of 12*(2x+7)*(x+1)^2 ?
The integral of 12*(2x+7)*(x+1)^2 is 12((x^4)/2+(11x^3)/3+8x^2+7x)+C