integral of (x^4-2x)^5(2x^3-1)

解

\int (x^{4} - 2 x)^{5} (2 x^{3} - 1) d x

\frac{x ^{24}}{12} - x^{21} + 5 x^{18} - \frac{40 x ^{15}}{3} + 20 x^{12} - 16 x^{9} + \frac{16 x ^{6}}{3} + C

解答ステップ

\int (x^{4} - 2 x)^{5} (2 x^{3} - 1) d x

拡張

(x^{4} - 2 x)^{5} (2 x^{3} - 1) : 2 x^{23} - 21 x^{20} + 90 x^{17} - 200 x^{14} + 240 x^{11} - 144 x^{8} + 32 x^{5}

= \int 2 x^{23} - 21 x^{20} + 90 x^{17} - 200 x^{14} + 240 x^{11} - 144 x^{8} + 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 2 x^{23} d x - \int 21 x^{20} d x + \int 90 x^{17} d x - \int 200 x^{14} d x + \int 240 x^{11} d x - \int 144 x^{8} d x + \int 32 x^{5} d x

\int 2 x^{23} d x = \frac{x ^{24}}{12}

\int 21 x^{20} d x = x^{21}

\int 90 x^{17} d x = 5 x^{18}

\int 200 x^{14} d x = \frac{40 x ^{15}}{3}

\int 240 x^{11} d x = 20 x^{12}

\int 144 x^{8} d x = 16 x^{9}

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

= \frac{x ^{24}}{12} - x^{21} + 5 x^{18} - \frac{40 x ^{15}}{3} + 20 x^{12} - 16 x^{9} + \frac{16 x ^{6}}{3}

定数を解答に追加する

= \frac{x ^{24}}{12} - x^{21} + 5 x^{18} - \frac{40 x ^{15}}{3} + 20 x^{12} - 16 x^{9} + \frac{16 x ^{6}}{3} + C