integral of (x-1/(x^2))(x^2+2/x)^7

解

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{7} d x

\frac{x ^{16}}{16} + x^{13} + 7 x^{10} + 28 x^{7} + 70 x^{4} + \frac{112}{x ^{2}} + \frac{64}{x ^{5}} + \frac{16}{x ^{8}} + 112 x + C

解答ステップ

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{7} d x

拡張

(x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{7} : x^{15} + 13 x^{12} + 70 x^{9} + 196 x^{6} + 280 x^{3} - \frac{224}{x ^{3}} - \frac{320}{x ^{6}} - \frac{128}{x ^{9}} + 112

= \int x^{15} + 13 x^{12} + 70 x^{9} + 196 x^{6} + 280 x^{3} - \frac{224}{x ^{3}} - \frac{320}{x ^{6}} - \frac{128}{x ^{9}} + 112 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^{15} d x + \int 13 x^{12} d x + \int 70 x^{9} d x + \int 196 x^{6} d x + \int 280 x^{3} d x - \int \frac{224}{x ^{3}} d x - \int \frac{320}{x ^{6}} d x - \int \frac{128}{x ^{9}} d x + \int 112 d x

\int x^{15} d x = \frac{x ^{16}}{16}

\int 13 x^{12} d x = x^{13}

\int 70 x^{9} d x = 7 x^{10}

\int 196 x^{6} d x = 28 x^{7}

\int 280 x^{3} d x = 70 x^{4}

\int \frac{224}{x ^{3}} d x = - \frac{112}{x ^{2}}

\int \frac{320}{x ^{6}} d x = - \frac{64}{x ^{5}}

\int \frac{128}{x ^{9}} d x = - \frac{16}{x ^{8}}

\int 112 d x = 112 x

= \frac{x ^{16}}{16} + x^{13} + 7 x^{10} + 28 x^{7} + 70 x^{4} - (- \frac{112}{x ^{2}}) - (- \frac{64}{x ^{5}}) - (- \frac{16}{x ^{8}}) + 112 x

簡素化

= \frac{x ^{16}}{16} + x^{13} + 7 x^{10} + 28 x^{7} + 70 x^{4} + \frac{112}{x ^{2}} + \frac{64}{x ^{5}} + \frac{16}{x ^{8}} + 112 x

定数を解答に追加する

= \frac{x ^{16}}{16} + x^{13} + 7 x^{10} + 28 x^{7} + 70 x^{4} + \frac{112}{x ^{2}} + \frac{64}{x ^{5}} + \frac{16}{x ^{8}} + 112 x + C