integral of (2-4x^2)/(x-x^3)

解

\int \frac{2 - 4 x ^{2}}{x - x ^{3}} d x

2 ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

解答ステップ

\int \frac{2 - 4 x ^{2}}{x - x ^{3}} d x

拡張

\frac{2 - 4 x ^{2}}{x - x ^{3}} : \frac{2}{x - x ^{3}} - \frac{4 x ^{2}}{x - x ^{3}}

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{2}{x - x ^{3}} d x - \int \frac{4 x ^{2}}{x - x ^{3}} d x

\int \frac{2}{x - x ^{3}} d x = 2 ln ∣ x ∣ - ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

\int \frac{4 x ^{2}}{x - x ^{3}} d x = 4 (- \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= 2 ln ∣ x ∣ - ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 4 (- \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

簡素化

2 ln ∣ x ∣ - ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ - 4 (- \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) : 2 ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

= 2 ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

定数を解答に追加する

= 2 ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C