What is the integral of (x+1)(9x-8) ?

The integral of (x+1)(9x-8) is (9x-8)((x^2)/2+x)-9((x^3)/6+(x^2)/2)+C

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integral of (x+1)(9x-8)

\int (x + 1) (9 x - 8) d x

(9 x - 8) (\frac{x ^{2}}{2} + x) - 9 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2}) + C

Solution steps

\int (x + 1) (9 x - 8) d x

Apply Integration By Parts

: (9 x - 8) (\frac{x ^{2}}{2} + x) - \int 9 (\frac{x ^{2}}{2} + x) d x

= (9 x - 8) (\frac{x ^{2}}{2} + x) - \int 9 (\frac{x ^{2}}{2} + x) d x

\int 9 (\frac{x ^{2}}{2} + x) d x = 9 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2})

= (9 x - 8) (\frac{x ^{2}}{2} + x) - 9 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2})

Add a constant to the solution

= (9 x - 8) (\frac{x ^{2}}{2} + x) - 9 (\frac{x ^{3}}{6} + \frac{x ^{2}}{2}) + C

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What is the integral of (x+1)(9x-8) ?
The integral of (x+1)(9x-8) is (9x-8)((x^2)/2+x)-9((x^3)/6+(x^2)/2)+C