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

解

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

6 (9 x^{3} - 9 x^{6} + 4 x^{9} - \frac{2 x ^{12}}{3}) + C

解答ステップ

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

定数を除く:

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

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

拡張

x^{2} (3 - 2 x^{3})^{3} : 27 x^{2} - 54 x^{5} + 36 x^{8} - 8 x^{11}

= 6 \cdot \int 27 x^{2} - 54 x^{5} + 36 x^{8} - 8 x^{11} 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

= 6 (\int 27 x^{2} d x - \int 54 x^{5} d x + \int 36 x^{8} d x - \int 8 x^{11} d x)

\int 27 x^{2} d x = 9 x^{3}

\int 54 x^{5} d x = 9 x^{6}

\int 36 x^{8} d x = 4 x^{9}

\int 8 x^{11} d x = \frac{2 x ^{12}}{3}

= 6 (9 x^{3} - 9 x^{6} + 4 x^{9} - \frac{2 x ^{12}}{3})

定数を解答に追加する

= 6 (9 x^{3} - 9 x^{6} + 4 x^{9} - \frac{2 x ^{12}}{3}) + C