integral of (x-1/(x^2))(x^2+2/x)^5

解

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{5} d x

\frac{x ^{12}}{12} + x^{9} + 5 x^{6} + \frac{40 x ^{3}}{3} + \frac{16}{x ^{3}} + \frac{16}{3 x ^{6}} + C

解答ステップ

\int (x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{5} d x

拡張

(x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{5} : x^{11} + 9 x^{8} + 30 x^{5} + 40 x^{2} - \frac{48}{x ^{4}} - \frac{32}{x ^{7}}

= \int x^{11} + 9 x^{8} + 30 x^{5} + 40 x^{2} - \frac{48}{x ^{4}} - \frac{32}{x ^{7}} 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^{11} d x + \int 9 x^{8} d x + \int 30 x^{5} d x + \int 40 x^{2} d x - \int \frac{48}{x ^{4}} d x - \int \frac{32}{x ^{7}} d x

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

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

\int 30 x^{5} d x = 5 x^{6}

\int 40 x^{2} d x = \frac{40 x ^{3}}{3}

\int \frac{48}{x ^{4}} d x = - \frac{16}{x ^{3}}

\int \frac{32}{x ^{7}} d x = - \frac{16}{3 x ^{6}}

= \frac{x ^{12}}{12} + x^{9} + 5 x^{6} + \frac{40 x ^{3}}{3} - (- \frac{16}{x ^{3}}) - (- \frac{16}{3 x ^{6}})

簡素化

= \frac{x ^{12}}{12} + x^{9} + 5 x^{6} + \frac{40 x ^{3}}{3} + \frac{16}{x ^{3}} + \frac{16}{3 x ^{6}}

定数を解答に追加する

= \frac{x ^{12}}{12} + x^{9} + 5 x^{6} + \frac{40 x ^{3}}{3} + \frac{16}{x ^{3}} + \frac{16}{3 x ^{6}} + C