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

解

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

- \frac{1}{x} - x + C

解答ステップ

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

拡張

\frac{1 - x ^{2}}{x ^{2}} : \frac{1}{x ^{2}} - 1

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

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

\int 1 d x = x

= - \frac{1}{x} - x

定数を解答に追加する

= - \frac{1}{x} - x + C