What is the integral of (9+u+u^2) ?

The integral of (9+u+u^2) is 9u+(u^2)/2+(u^3)/3+C

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integral of (9+u+u^2)

\int (9 + u + u^{2}) d u

9 u + \frac{u ^{2}}{2} + \frac{u ^{3}}{3} + C

Solution steps

\int 9 + u + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 d u + \int u d u + \int u^{2} d u

\int 9 d u = 9 u

\int u d u = \frac{u ^{2}}{2}

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

= 9 u + \frac{u ^{2}}{2} + \frac{u ^{3}}{3}

Add a constant to the solution

= 9 u + \frac{u ^{2}}{2} + \frac{u ^{3}}{3} + C

\int \frac{4}{x ^{2} ( x - 2 )} d x

\int (x^{4} + x + 9) d x

x \to - 1 lim (3 x - 5)

ln (\frac{x ^{2} - 1}{x ^{2} + 1})

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

What is the integral of (9+u+u^2) ?
The integral of (9+u+u^2) is 9u+(u^2)/2+(u^3)/3+C