What is the integral of (x^2+18)(x^5-9) ?

The integral of (x^2+18)(x^5-9) is (x^8)/8-3x^3+3x^6-162x+C

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integral of (x^2+18)(x^5-9)

\int (x^{2} + 18) (x^{5} - 9) d x

\frac{x ^{8}}{8} - 3 x^{3} + 3 x^{6} - 162 x + C

Solution steps

\int (x^{2} + 18) (x^{5} - 9) d x

Expand

(x^{2} + 18) (x^{5} - 9) : x^{7} - 9 x^{2} + 18 x^{5} - 162

= \int x^{7} - 9 x^{2} + 18 x^{5} - 162 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 x^{7} d x - \int 9 x^{2} d x + \int 18 x^{5} d x - \int 162 d x

\int x^{7} d x = \frac{x ^{8}}{8}

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

\int 18 x^{5} d x = 3 x^{6}

\int 162 d x = 162 x

= \frac{x ^{8}}{8} - 3 x^{3} + 3 x^{6} - 162 x

Add a constant to the solution

= \frac{x ^{8}}{8} - 3 x^{3} + 3 x^{6} - 162 x + C

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What is the integral of (x^2+18)(x^5-9) ?
The integral of (x^2+18)(x^5-9) is (x^8)/8-3x^3+3x^6-162x+C