integral of (-6x^2+3x+5)/(x^3-x)

解答

\int \frac{- 6 x ^{2} + 3 x + 5}{x ^{3} - x} d x

ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - 3 ln x^{2} - 1 - 5 ln ∣ x ∣ + C

求解步骤

\int \frac{- 6 x ^{2} + 3 x + 5}{x ^{3} - x} d x

乘开

\frac{- 6 x ^{2} + 3 x + 5}{x ^{3} - x} : - \frac{6 x ^{2}}{x ^{3} - x} + \frac{3 x}{x ^{3} - x} + \frac{5}{x ^{3} - x}

使用积分加法定则:

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

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

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

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

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

化简

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

= ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - 3 ln x^{2} - 1 - 5 ln ∣ x ∣

解答补常数

= ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - 3 ln x^{2} - 1 - 5 ln ∣ x ∣ + C

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