What is the integral of (9x^2-3x+5) ?

The integral of (9x^2-3x+5) is 3x^3-(3x^2)/2+5x+C

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integral of (9x^2-3x+5)

\int (9 x^{2} - 3 x + 5) d x

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

Solution steps

\int 9 x^{2} - 3 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 9 x^{2} d x - \int 3 x d x + \int 5 d x

\int 9 x^{2} d x = 3 x^{3}

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

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What is the integral of (9x^2-3x+5) ?
The integral of (9x^2-3x+5) is 3x^3-(3x^2)/2+5x+C