integral of (x^2-5x+2)/(x^3-x^2-x+1)

解

\int \frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} d x

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

解答ステップ

\int \frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} d x

拡張

\frac{x ^{2} - 5 x + 2}{x ^{3} - x ^{2} - x + 1} : \frac{x ^{2}}{x ^{3} - x ^{2} - x + 1} - \frac{5 x}{x ^{3} - x ^{2} - x + 1} + \frac{2}{x ^{3} - x ^{2} - x + 1}

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{x ^{2}}{x ^{3} - x ^{2} - x + 1} d x - \int \frac{5 x}{x ^{3} - x ^{2} - x + 1} d x + \int \frac{2}{x ^{3} - x ^{2} - x + 1} d x

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

\int \frac{5 x}{x ^{3} - x ^{2} - x + 1} d x = 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣)

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

= \frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣) - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{x - 1} + \frac{1}{2} ln ∣ x + 1 ∣

簡素化

\frac{3}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣) - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{x - 1} + \frac{1}{2} ln ∣ x + 1 ∣ : - ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ + \frac{1}{x - 1}

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

定数を解答に追加する

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