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

解

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

\frac{27 x ^{15}}{5} - 18 x^{12} + 24 x^{9} - 16 x^{6} + \frac{16 x ^{3}}{3} + C

解答ステップ

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

拡張

x^{2} (3 x^{3} - 2)^{4} : 81 x^{14} - 216 x^{11} + 216 x^{8} - 96 x^{5} + 16 x^{2}

= \int 81 x^{14} - 216 x^{11} + 216 x^{8} - 96 x^{5} + 16 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

= \int 81 x^{14} d x - \int 216 x^{11} d x + \int 216 x^{8} d x - \int 96 x^{5} d x + \int 16 x^{2} d x

\int 81 x^{14} d x = \frac{27 x ^{15}}{5}

\int 216 x^{11} d x = 18 x^{12}

\int 216 x^{8} d x = 24 x^{9}

\int 96 x^{5} d x = 16 x^{6}

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

= \frac{27 x ^{15}}{5} - 18 x^{12} + 24 x^{9} - 16 x^{6} + \frac{16 x ^{3}}{3}

定数を解答に追加する

= \frac{27 x ^{15}}{5} - 18 x^{12} + 24 x^{9} - 16 x^{6} + \frac{16 x ^{3}}{3} + C