What is the integral of (7^3+6x^2-9x-8) ?

The integral of (7^3+6x^2-9x-8) is 2x^3-(9x^2)/2+335x+C

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integral of (7^3+6x^2-9x-8)

\int (7^{3} + 6 x^{2} - 9 x - 8) d x

2 x^{3} - \frac{9 x ^{2}}{2} + 335 x + C

Solution steps

\int 7^{3} + 6 x^{2} - 9 x - 8 d x

Simplify

= \int 6 x^{2} - 9 x + 335 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 6 x^{2} d x - \int 9 x d x + \int 335 d x

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

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

\int 335 d x = 335 x

= 2 x^{3} - \frac{9 x ^{2}}{2} + 335 x

Add a constant to the solution

= 2 x^{3} - \frac{9 x ^{2}}{2} + 335 x + C

What is the integral of (7^3+6x^2-9x-8) ?
The integral of (7^3+6x^2-9x-8) is 2x^3-(9x^2)/2+335x+C