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

解

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

ln ∣ x - 2 ∣ + \frac{1}{2} ln 4 x^{2} + 4 x + 4 - 3 arctan (\frac{1}{3} (2 x + 1)) + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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