integral of (x-1/x)^2

解

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

\frac{x ^{3}}{3} - 2 x - \frac{1}{x} + C

解答ステップ

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

拡張

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

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

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

\int 2 d x = 2 x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= \frac{x ^{3}}{3} - 2 x - \frac{1}{x}

定数を解答に追加する

= \frac{x ^{3}}{3} - 2 x - \frac{1}{x} + C