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

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

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integral of x(x-2)^2

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

\frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 2 x^{2} + C

Solution steps

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

Expand

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

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

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 2 x^{2}

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 2 x^{2} + C

\frac{d}{d x} (cos^{2} (- 3 x^{2} + 2))

f (x) = x^{2} - 4

- 2 y^{^{''}} + 9 y = 0, y (- 2) = 2, y^{^{'}} (2) = 1

\frac{d}{d x} (in (x + 1 + x^{2}))

f (x) = 2 x^{3} - x^{2} + 1, (1, 2)

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