integral of (x^2+2x-6)/(x^3-x)

解

\int \frac{x ^{2} + 2 x - 6}{x ^{3} - x} d x

6 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 1 ∣ + C

解答ステップ

\int \frac{x ^{2} + 2 x - 6}{x ^{3} - x} d x

以下の部分分数を得る:

\frac{x ^{2} + 2 x - 6}{x ^{3} - x} : \frac{6}{x} - \frac{7}{2 ( x + 1 )} - \frac{3}{2 ( x - 1 )}

= \int \frac{6}{x} - \frac{7}{2 ( x + 1 )} - \frac{3}{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{6}{x} d x - \int \frac{7}{2 ( x + 1 )} d x - \int \frac{3}{2 ( x - 1 )} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{7}{2 ( x + 1 )} d x = \frac{7}{2} ln ∣ x + 1 ∣

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

= 6 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 1 ∣

定数を解答に追加する

= 6 ln ∣ x ∣ - \frac{7}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 1 ∣ + C