integral of (x/3-1)^9

解

\int (\frac{x}{3} - 1)^{9} d x

\frac{x ^{10}}{196830} - \frac{x ^{9}}{6561} + \frac{x ^{8}}{486} - \frac{4 x ^{7}}{243} + \frac{7 x ^{6}}{81} - \frac{14 x ^{5}}{45} + \frac{7 x ^{4}}{9} - \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} - x + C

解答ステップ

\int (\frac{x}{3} - 1)^{9} d x

拡張

(\frac{x}{3} - 1)^{9} : \frac{x ^{9}}{19683} - \frac{x ^{8}}{729} + \frac{4 x ^{7}}{243} - \frac{28 x ^{6}}{243} + \frac{14 x ^{5}}{27} - \frac{14 x ^{4}}{9} + \frac{28 x ^{3}}{9} - 4 x^{2} + 3 x - 1

= \int \frac{x ^{9}}{19683} - \frac{x ^{8}}{729} + \frac{4 x ^{7}}{243} - \frac{28 x ^{6}}{243} + \frac{14 x ^{5}}{27} - \frac{14 x ^{4}}{9} + \frac{28 x ^{3}}{9} - 4 x^{2} + 3 x - 1 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 \frac{x ^{9}}{19683} d x - \int \frac{x ^{8}}{729} d x + \int \frac{4 x ^{7}}{243} d x - \int \frac{28 x ^{6}}{243} d x + \int \frac{14 x ^{5}}{27} d x - \int \frac{14 x ^{4}}{9} d x + \int \frac{28 x ^{3}}{9} d x - \int 4 x^{2} d x + \int 3 x d x - \int 1 d x

\int \frac{x ^{9}}{19683} d x = \frac{x ^{10}}{196830}

\int \frac{x ^{8}}{729} d x = \frac{x ^{9}}{6561}

\int \frac{4 x ^{7}}{243} d x = \frac{x ^{8}}{486}

\int \frac{28 x ^{6}}{243} d x = \frac{4 x ^{7}}{243}

\int \frac{14 x ^{5}}{27} d x = \frac{7 x ^{6}}{81}

\int \frac{14 x ^{4}}{9} d x = \frac{14 x ^{5}}{45}

\int \frac{28 x ^{3}}{9} d x = \frac{7 x ^{4}}{9}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

\int 1 d x = x

= \frac{x ^{10}}{196830} - \frac{x ^{9}}{6561} + \frac{x ^{8}}{486} - \frac{4 x ^{7}}{243} + \frac{7 x ^{6}}{81} - \frac{14 x ^{5}}{45} + \frac{7 x ^{4}}{9} - \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} - x

定数を解答に追加する

= \frac{x ^{10}}{196830} - \frac{x ^{9}}{6561} + \frac{x ^{8}}{486} - \frac{4 x ^{7}}{243} + \frac{7 x ^{6}}{81} - \frac{14 x ^{5}}{45} + \frac{7 x ^{4}}{9} - \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2} - x + C