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

解

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

- \frac{1}{x} - 2 ln ∣ 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}} - \frac{2}{x} + 1

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 1 d x = x

= - \frac{1}{x} - 2 ln ∣ x ∣ + x

定数を解答に追加する

= - \frac{1}{x} - 2 ln ∣ x ∣ + x + C