integral of (x^2+5x-4)/(x^3-x)

解答

\int \frac{x ^{2} + 5 x - 4}{x ^{3} - x} d x

\frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

求解步骤

\int \frac{x ^{2} + 5 x - 4}{x ^{3} - x} d x

乘开

\frac{x ^{2} + 5 x - 4}{x ^{3} - x} : \frac{x ^{2}}{x ^{3} - x} + \frac{5 x}{x ^{3} - x} - \frac{4}{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{x ^{2}}{x ^{3} - x} d x + \int \frac{5 x}{x ^{3} - x} d x - \int \frac{4}{x ^{3} - x} d x

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

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

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

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

化简

\frac{1}{2} ln x^{2} - 1 - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) - 4 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) : \frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

= \frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

解答补常数

= \frac{1}{2} ln x^{2} - 1 - \frac{9}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

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