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

解

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

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

解答ステップ

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

拡張

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

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

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

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

= \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - (- \frac{1}{2} ln 1 - x^{2})

簡素化

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

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

定数を解答に追加する

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