integral of (x^2-2x)^5(x-1)

解

\int (x^{2} - 2 x)^{5} (x - 1) d x

\frac{x ^{12}}{12} - x^{11} + 5 x^{10} - \frac{40 x ^{9}}{3} + 20 x^{8} - 16 x^{7} + \frac{16 x ^{6}}{3} + C

解答ステップ

\int (x^{2} - 2 x)^{5} (x - 1) d x

拡張

(x^{2} - 2 x)^{5} (x - 1) : x^{11} - 11 x^{10} + 50 x^{9} - 120 x^{8} + 160 x^{7} - 112 x^{6} + 32 x^{5}

= \int x^{11} - 11 x^{10} + 50 x^{9} - 120 x^{8} + 160 x^{7} - 112 x^{6} + 32 x^{5} 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^{11} d x - \int 11 x^{10} d x + \int 50 x^{9} d x - \int 120 x^{8} d x + \int 160 x^{7} d x - \int 112 x^{6} d x + \int 32 x^{5} d x

\int x^{11} d x = \frac{x ^{12}}{12}

\int 11 x^{10} d x = x^{11}

\int 50 x^{9} d x = 5 x^{10}

\int 120 x^{8} d x = \frac{40 x ^{9}}{3}

\int 160 x^{7} d x = 20 x^{8}

\int 112 x^{6} d x = 16 x^{7}

\int 32 x^{5} d x = \frac{16 x ^{6}}{3}

= \frac{x ^{12}}{12} - x^{11} + 5 x^{10} - \frac{40 x ^{9}}{3} + 20 x^{8} - 16 x^{7} + \frac{16 x ^{6}}{3}

定数を解答に追加する

= \frac{x ^{12}}{12} - x^{11} + 5 x^{10} - \frac{40 x ^{9}}{3} + 20 x^{8} - 16 x^{7} + \frac{16 x ^{6}}{3} + C