What is the integral of 1+3x+18x^2 ?

The integral of 1+3x+18x^2 is x+(3x^2)/2+6x^3+C

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integral of 1+3x+18x^2

\int 1 + 3 x + 18 x^{2} d x

x + \frac{3 x ^{2}}{2} + 6 x^{3} + C

Solution steps

\int 1 + 3 x + 18 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

= \int 1 d x + \int 3 x d x + \int 18 x^{2} d x

\int 1 d x = x

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

\int 18 x^{2} d x = 6 x^{3}

= x + \frac{3 x ^{2}}{2} + 6 x^{3}

Add a constant to the solution

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

x \to 0 + lim (\frac{ln ( x ) + 1}{x})

\int (4 sin (x))^{3} d x

\frac{3 ( s + 1 )}{s ^{2} + \frac{1}{2}}

e^{- \frac{x}{2}}, 0

n = 2 \sum \infty (\frac{1}{2})^{3 n}

What is the integral of 1+3x+18x^2 ?
The integral of 1+3x+18x^2 is x+(3x^2)/2+6x^3+C