What is the integral of (y-1)(2y+1) ?

The integral of (y-1)(2y+1) is (2y^3)/3-(y^2)/2-y+C

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integral of (y-1)(2y+1)

\int (y - 1) (2 y + 1) d y

\frac{2 y ^{3}}{3} - \frac{y ^{2}}{2} - y + C

Solution steps

\int (y - 1) (2 y + 1) d y

Expand

(y - 1) (2 y + 1) : 2 y^{2} - y - 1

= \int 2 y^{2} - y - 1 d y

Apply the Sum Rule:

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

= \int 2 y^{2} d y - \int y d y - \int 1 d y

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

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

\int 1 d y = y

= \frac{2 y ^{3}}{3} - \frac{y ^{2}}{2} - y

Add a constant to the solution

= \frac{2 y ^{3}}{3} - \frac{y ^{2}}{2} - y + C

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What is the integral of (y-1)(2y+1) ?
The integral of (y-1)(2y+1) is (2y^3)/3-(y^2)/2-y+C