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

解

\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

解答ステップ

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

拡張

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

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

総和規則を適用する:

\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

定数を解答に追加する

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