What is the integral of 2pi(4x^2+4x+1) ?

The integral of 2pi(4x^2+4x+1) is 2pi((4x^3)/3+2x^2+x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2pi(4x^2+4x+1)

\int 2 π (4 x^{2} + 4 x + 1) d x

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

Solution steps

\int 2 π (4 x^{2} + 4 x + 1) d x

Take the constant out:

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

= 2 π \cdot \int 4 x^{2} + 4 x + 1 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

= 2 π (\int 4 x^{2} d x + \int 4 x d x + \int 1 d x)

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

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

\int 1 d x = x

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

Add a constant to the solution

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

What is the integral of 2pi(4x^2+4x+1) ?
The integral of 2pi(4x^2+4x+1) is 2pi((4x^3)/3+2x^2+x)+C