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

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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

\frac{d}{d x} (csc^{13} (x))

x \to \frac{2}{3} lim (\frac{27 x ^{3} - 8}{9 x ^{2} - 4})

t y^{^{'''}} - y^{^{''}} = 0

8 - x

f (x) = \frac{cos ( x )}{x ^{3}}

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