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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

\int 1 d x = x

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

定数を解答に追加する

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