What is the integral of (3x^3+6x^2-12x+5) ?

The integral of (3x^3+6x^2-12x+5) is (3x^4)/4+2x^3-6x^2+5x+C

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integral of (3x^3+6x^2-12x+5)

\int (3 x^{3} + 6 x^{2} - 12 x + 5) d x

\frac{3 x ^{4}}{4} + 2 x^{3} - 6 x^{2} + 5 x + C

Solution steps

\int 3 x^{3} + 6 x^{2} - 12 x + 5 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 3 x^{3} d x + \int 6 x^{2} d x - \int 12 x d x + \int 5 d x

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

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

\int 12 x d x = 6 x^{2}

\int 5 d x = 5 x

= \frac{3 x ^{4}}{4} + 2 x^{3} - 6 x^{2} + 5 x

Add a constant to the solution

= \frac{3 x ^{4}}{4} + 2 x^{3} - 6 x^{2} + 5 x + C

\frac{d}{d x} (\frac{9}{1 + 9 x})

f (x) = t^{5} sin (t)

\int_{0}^{1} 1 + 144 x d x

\frac{d y}{d x} = \frac{x ( y - 2 )}{x ^{2} + 4}

t^{2} t

What is the integral of (3x^3+6x^2-12x+5) ?
The integral of (3x^3+6x^2-12x+5) is (3x^4)/4+2x^3-6x^2+5x+C