integral of (2x^3-x)^4(6x^2-1)

解

\int (2 x^{3} - x)^{4} (6 x^{2} - 1) d x

\frac{32 x ^{15}}{5} - 16 x^{13} + 16 x^{11} - 8 x^{9} + 2 x^{7} - \frac{x ^{5}}{5} + C

解答ステップ

\int (2 x^{3} - x)^{4} (6 x^{2} - 1) d x

拡張

(2 x^{3} - x)^{4} (6 x^{2} - 1) : 96 x^{14} - 208 x^{12} + 176 x^{10} - 72 x^{8} + 14 x^{6} - x^{4}

= \int 96 x^{14} - 208 x^{12} + 176 x^{10} - 72 x^{8} + 14 x^{6} - x^{4} 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 96 x^{14} d x - \int 208 x^{12} d x + \int 176 x^{10} d x - \int 72 x^{8} d x + \int 14 x^{6} d x - \int x^{4} d x

\int 96 x^{14} d x = \frac{32 x ^{15}}{5}

\int 208 x^{12} d x = 16 x^{13}

\int 176 x^{10} d x = 16 x^{11}

\int 72 x^{8} d x = 8 x^{9}

\int 14 x^{6} d x = 2 x^{7}

\int x^{4} d x = \frac{x ^{5}}{5}

= \frac{32 x ^{15}}{5} - 16 x^{13} + 16 x^{11} - 8 x^{9} + 2 x^{7} - \frac{x ^{5}}{5}

定数を解答に追加する

= \frac{32 x ^{15}}{5} - 16 x^{13} + 16 x^{11} - 8 x^{9} + 2 x^{7} - \frac{x ^{5}}{5} + C