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

The integral of (x+1)(x-2)^2 is (x^4)/4-x^3+4x+C

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integral of (x+1)(x-2)^2

\int (x + 1) (x - 2)^{2} d x

\frac{x ^{4}}{4} - x^{3} + 4 x + C

Solution steps

\int (x + 1) (x - 2)^{2} d x

x - 2 = 2 (\frac{x}{2} - 1)

= \int (x + 1) (2 (\frac{x}{2} - 1))^{2} d x

Simplify

(x + 1) (2 (\frac{x}{2} - 1))^{2} : x^{3} - 3 x^{2} + 4

= \int x^{3} - 3 x^{2} + 4 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^{3} d x - \int 3 x^{2} d x + \int 4 d x

\int x^{3} d x = \frac{x ^{4}}{4}

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

\int 4 d x = 4 x

= \frac{x ^{4}}{4} - x^{3} + 4 x

Add a constant to the solution

= \frac{x ^{4}}{4} - x^{3} + 4 x + C

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What is the integral of (x+1)(x-2)^2 ?
The integral of (x+1)(x-2)^2 is (x^4)/4-x^3+4x+C