integral of (x^3-1)/(x^2-1)

解答

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

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

求解步骤

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

乘开

\frac{x ^{3} - 1}{x ^{2} - 1} : \frac{x ^{3}}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

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

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

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

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

化简

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

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

解答补常数

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

\int 3600 - 120 x d x

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