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

The integral of (x^3)/3-(5x^2)/2+6x+C is (x^4}{12}-\frac{5x^3)/6+3x^2+Cx+C

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

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

\frac{x ^{4}}{12} - \frac{5 x ^{3}}{6} + 3 x^{2} + C x + C

Solution steps

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

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

\int \frac{5 x ^{2}}{2} d x = \frac{5 x ^{3}}{6}

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

\int C d x = C x

= \frac{x ^{4}}{12} - \frac{5 x ^{3}}{6} + 3 x^{2} + C x

Add a constant to the solution

= \frac{x ^{4}}{12} - \frac{5 x ^{3}}{6} + 3 x^{2} + C x + C

\int \frac{1}{200 + 3 t} d t

n = 3 \sum \infty \frac{4}{n ^{2} - 1}

x \to 0 lim (- 3 e^{3 x})

\int x e^{- \frac{x}{4}} d x

x + \frac{1}{x}, at x = 4

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