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

解答

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

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

求解步骤

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

将

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

用部份分式展开:

\frac{1}{x - 1} + \frac{- x - 1}{x ^{2} + x + 1}

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

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

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

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

解答补常数

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

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