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

解

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

- 2 ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + 3 ln x^{2} - 1 + 3 ln ∣ x ∣ + C

解答ステップ

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

拡張

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

\int \frac{6 x ^{2}}{x ^{3} - x} d x = 3 ln x^{2} - 1

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

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

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

簡素化

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

= - 2 ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + 3 ln x^{2} - 1 + 3 ln ∣ x ∣

定数を解答に追加する

= - 2 ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + 3 ln x^{2} - 1 + 3 ln ∣ x ∣ + C