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

The integral of (x-1)^2(x-2) is (x^4)/4-(4x^3)/3+(5x^2)/2-2x+C

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

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

\frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{5 x ^{2}}{2} - 2 x + C

Solution steps

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

Expand

(x - 1)^{2} (x - 2) : x^{3} - 4 x^{2} + 5 x - 2

= \int x^{3} - 4 x^{2} + 5 x - 2 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 5 x d x - \int 2 d x

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

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 2 d x = 2 x

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{5 x ^{2}}{2} - 2 x

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{5 x ^{2}}{2} - 2 x + C

\int \frac{x ^{3} + 1}{x x} d x

\int \frac{e ^{\frac{y}{x}}}{x} d x

\int 3 x (5 x^{2} - 3 x + 2) d x

\int 5 x^{\frac{3}{2}} + 2 x - 7 x + 3 d x

\int \frac{e ^{x} + e ^{2 x} + e ^{3 x}}{e ^{4 x}} d x

What is the integral of (x-1)^2(x-2) ?
The integral of (x-1)^2(x-2) is (x^4)/4-(4x^3)/3+(5x^2)/2-2x+C