integral of (x^4-9x)^2(4x^3-9)

解

\int (x^{4} - 9 x)^{2} (4 x^{3} - 9) d x

\frac{x ^{12}}{3} - 9 x^{9} + 81 x^{6} - 243 x^{3} + C

解答ステップ

\int (x^{4} - 9 x)^{2} (4 x^{3} - 9) d x

拡張

(x^{4} - 9 x)^{2} (4 x^{3} - 9) : 4 x^{11} - 81 x^{8} + 486 x^{5} - 729 x^{2}

= \int 4 x^{11} - 81 x^{8} + 486 x^{5} - 729 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 4 x^{11} d x - \int 81 x^{8} d x + \int 486 x^{5} d x - \int 729 x^{2} d x

\int 4 x^{11} d x = \frac{x ^{12}}{3}

\int 81 x^{8} d x = 9 x^{9}

\int 486 x^{5} d x = 81 x^{6}

\int 729 x^{2} d x = 243 x^{3}

= \frac{x ^{12}}{3} - 9 x^{9} + 81 x^{6} - 243 x^{3}

定数を解答に追加する

= \frac{x ^{12}}{3} - 9 x^{9} + 81 x^{6} - 243 x^{3} + C