integral of ((x+1)^2(3-x)^6)

解

\int ((x + 1)^{2} (3 - x)^{6}) d x

\frac{x ^{9}}{9} - 2 x^{8} + \frac{100 x ^{7}}{7} - 48 x^{6} + 54 x^{5} + 108 x^{4} - 324 x^{3} + 729 x + C

解答ステップ

\int (x + 1)^{2} (3 - x)^{6} d x

拡張

(x + 1)^{2} (3 - x)^{6} : x^{8} - 16 x^{7} + 100 x^{6} - 288 x^{5} + 270 x^{4} + 432 x^{3} - 972 x^{2} + 729

= \int x^{8} - 16 x^{7} + 100 x^{6} - 288 x^{5} + 270 x^{4} + 432 x^{3} - 972 x^{2} + 729 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 x^{8} d x - \int 16 x^{7} d x + \int 100 x^{6} d x - \int 288 x^{5} d x + \int 270 x^{4} d x + \int 432 x^{3} d x - \int 972 x^{2} d x + \int 729 d x

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

\int 16 x^{7} d x = 2 x^{8}

\int 100 x^{6} d x = \frac{100 x ^{7}}{7}

\int 288 x^{5} d x = 48 x^{6}

\int 270 x^{4} d x = 54 x^{5}

\int 432 x^{3} d x = 108 x^{4}

\int 972 x^{2} d x = 324 x^{3}

\int 729 d x = 729 x

= \frac{x ^{9}}{9} - 2 x^{8} + \frac{100 x ^{7}}{7} - 48 x^{6} + 54 x^{5} + 108 x^{4} - 324 x^{3} + 729 x

定数を解答に追加する

= \frac{x ^{9}}{9} - 2 x^{8} + \frac{100 x ^{7}}{7} - 48 x^{6} + 54 x^{5} + 108 x^{4} - 324 x^{3} + 729 x + C