integral of-4x^2(4x^3-1)^7

解

\int - 4 x^{2} (4 x^{3} - 1)^{7} d x

- 4 (\frac{2048 x ^{24}}{3} - \frac{4096 x ^{21}}{3} + \frac{3584 x ^{18}}{3} - \frac{1792 x ^{15}}{3} + \frac{560 x ^{12}}{3} - \frac{112 x ^{9}}{3} + \frac{14 x ^{6}}{3} - \frac{x ^{3}}{3}) + C

解答ステップ

\int - 4 x^{2} (4 x^{3} - 1)^{7} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 4 \cdot \int x^{2} (4 x^{3} - 1)^{7} d x

拡張

x^{2} (4 x^{3} - 1)^{7} : 16384 x^{23} - 28672 x^{20} + 21504 x^{17} - 8960 x^{14} + 2240 x^{11} - 336 x^{8} + 28 x^{5} - x^{2}

= - 4 \cdot \int 16384 x^{23} - 28672 x^{20} + 21504 x^{17} - 8960 x^{14} + 2240 x^{11} - 336 x^{8} + 28 x^{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

= - 4 (\int 16384 x^{23} d x - \int 28672 x^{20} d x + \int 21504 x^{17} d x - \int 8960 x^{14} d x + \int 2240 x^{11} d x - \int 336 x^{8} d x + \int 28 x^{5} d x - \int x^{2} d x)

\int 16384 x^{23} d x = \frac{2048 x ^{24}}{3}

\int 28672 x^{20} d x = \frac{4096 x ^{21}}{3}

\int 21504 x^{17} d x = \frac{3584 x ^{18}}{3}

\int 8960 x^{14} d x = \frac{1792 x ^{15}}{3}

\int 2240 x^{11} d x = \frac{560 x ^{12}}{3}

\int 336 x^{8} d x = \frac{112 x ^{9}}{3}

\int 28 x^{5} d x = \frac{14 x ^{6}}{3}

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

= - 4 (\frac{2048 x ^{24}}{3} - \frac{4096 x ^{21}}{3} + \frac{3584 x ^{18}}{3} - \frac{1792 x ^{15}}{3} + \frac{560 x ^{12}}{3} - \frac{112 x ^{9}}{3} + \frac{14 x ^{6}}{3} - \frac{x ^{3}}{3})

定数を解答に追加する

= - 4 (\frac{2048 x ^{24}}{3} - \frac{4096 x ^{21}}{3} + \frac{3584 x ^{18}}{3} - \frac{1792 x ^{15}}{3} + \frac{560 x ^{12}}{3} - \frac{112 x ^{9}}{3} + \frac{14 x ^{6}}{3} - \frac{x ^{3}}{3}) + C