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

解

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

- ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{10}{3 3} arctan (\frac{1}{3} (2 x - 1)) - \frac{2}{3 3} sin (2 arctan (\frac{1}{3} (2 x - 1))) + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

\int \frac{1}{( x ^{2} - x + 1 ) ^{2}} d x = \frac{2}{3 3} (2 arctan (\frac{2 x - 1}{3}) + sin (2 arctan (\frac{2 x - 1}{3})))

= - ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{2}{3} arctan (\frac{2 x - 1}{3}) - \frac{2}{3 3} (2 arctan (\frac{2 x - 1}{3}) + sin (2 arctan (\frac{2 x - 1}{3})))

簡素化

- ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{2}{3} arctan (\frac{2 x - 1}{3}) - \frac{2}{3 3} (2 arctan (\frac{2 x - 1}{3}) + sin (2 arctan (\frac{2 x - 1}{3}))) : - ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{10}{3 3} arctan (\frac{1}{3} (2 x - 1)) - \frac{2}{3 3} sin (2 arctan (\frac{1}{3} (2 x - 1)))

= - ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{10}{3 3} arctan (\frac{1}{3} (2 x - 1)) - \frac{2}{3 3} sin (2 arctan (\frac{1}{3} (2 x - 1)))

定数を解答に追加する

= - ln ∣ x ∣ + ln ∣ x - 1 ∣ - \frac{10}{3 3} arctan (\frac{1}{3} (2 x - 1)) - \frac{2}{3 3} sin (2 arctan (\frac{1}{3} (2 x - 1))) + C