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

解

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

\frac{1}{8} ln ∣ x - 3 ∣ - \frac{3}{8} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x - 2}{x ^{3} - 3 x ^{2} - x + 3} : \frac{1}{8 ( x - 3 )} - \frac{3}{8 ( x + 1 )} + \frac{1}{4 ( x - 1 )}

= \int \frac{1}{8 ( x - 3 )} - \frac{3}{8 ( x + 1 )} + \frac{1}{4 ( 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}{8 ( x - 3 )} d x - \int \frac{3}{8 ( x + 1 )} d x + \int \frac{1}{4 ( x - 1 )} d x

\int \frac{1}{8 ( x - 3 )} d x = \frac{1}{8} ln ∣ x - 3 ∣

\int \frac{3}{8 ( x + 1 )} d x = \frac{3}{8} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{3}{8} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣

定数を解答に追加する

= \frac{1}{8} ln ∣ x - 3 ∣ - \frac{3}{8} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣ + C