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

解

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

2 (\frac{729 x ^{8}}{8} - \frac{2916 x ^{7}}{7} + 810 x^{6} - 864 x^{5} + 540 x^{4} - 192 x^{3} + 32 x^{2}) + C

解答ステップ

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

定数を除く:

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

= 2 \cdot \int x (3 x - 2)^{6} d x

拡張

x (3 x - 2)^{6} : 729 x^{7} - 2916 x^{6} + 4860 x^{5} - 4320 x^{4} + 2160 x^{3} - 576 x^{2} + 64 x

= 2 \cdot \int 729 x^{7} - 2916 x^{6} + 4860 x^{5} - 4320 x^{4} + 2160 x^{3} - 576 x^{2} + 64 x 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

= 2 (\int 729 x^{7} d x - \int 2916 x^{6} d x + \int 4860 x^{5} d x - \int 4320 x^{4} d x + \int 2160 x^{3} d x - \int 576 x^{2} d x + \int 64 x d x)

\int 729 x^{7} d x = \frac{729 x ^{8}}{8}

\int 2916 x^{6} d x = \frac{2916 x ^{7}}{7}

\int 4860 x^{5} d x = 810 x^{6}

\int 4320 x^{4} d x = 864 x^{5}

\int 2160 x^{3} d x = 540 x^{4}

\int 576 x^{2} d x = 192 x^{3}

\int 64 x d x = 32 x^{2}

= 2 (\frac{729 x ^{8}}{8} - \frac{2916 x ^{7}}{7} + 810 x^{6} - 864 x^{5} + 540 x^{4} - 192 x^{3} + 32 x^{2})

定数を解答に追加する

= 2 (\frac{729 x ^{8}}{8} - \frac{2916 x ^{7}}{7} + 810 x^{6} - 864 x^{5} + 540 x^{4} - 192 x^{3} + 32 x^{2}) + C