integral of (-2x+4)/(x^4-2x^3+2x^2-2x+1)

解

\int \frac{- 2 x + 4}{x ^{4} - 2 x ^{3} + 2 x ^{2} - 2 x + 1} d x

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

解答ステップ

\int \frac{- 2 x + 4}{x ^{4} - 2 x ^{3} + 2 x ^{2} - 2 x + 1} d x

以下の部分分数を得る:

\frac{- 2 x + 4}{x ^{4} - 2 x ^{3} + 2 x ^{2} - 2 x + 1} : - \frac{2}{x - 1} + \frac{1}{( x - 1 ) ^{2}} + \frac{2 x + 1}{x ^{2} + 1}

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

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

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

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

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

定数を解答に追加する

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