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

解

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

\frac{x ^{6}}{6} + x^{3} + \frac{4}{3 x ^{3}} + C

解答ステップ

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

拡張

(x - \frac{1}{x ^{2}}) (x^{2} + \frac{2}{x})^{2} : x^{5} + 3 x^{2} - \frac{4}{x ^{4}}

= \int x^{5} + 3 x^{2} - \frac{4}{x ^{4}} 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^{5} d x + \int 3 x^{2} d x - \int \frac{4}{x ^{4}} d x

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

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

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

= \frac{x ^{6}}{6} + x^{3} - (- \frac{4}{3 x ^{3}})

簡素化

= \frac{x ^{6}}{6} + x^{3} + \frac{4}{3 x ^{3}}

定数を解答に追加する

= \frac{x ^{6}}{6} + x^{3} + \frac{4}{3 x ^{3}} + C