integral of 1/(z^3+1)

解答

\int \frac{1}{z ^{3} + 1} d z

\frac{1}{3} ln ∣ z + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 z ^{2} - 4 z}{3} + 23 arctan (\frac{2 z - 1}{3})) + C

求解步骤

\int \frac{1}{z ^{3} + 1} d z

将

\frac{1}{z ^{3} + 1}

用部份分式展开:

\frac{1}{3 ( z + 1 )} + \frac{- z + 2}{3 ( z ^{2} - z + 1 )}

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

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}{3 ( z + 1 )} d z + \int \frac{- z + 2}{3 ( z ^{2} - z + 1 )} d z

\int \frac{1}{3 ( z + 1 )} d z = \frac{1}{3} ln ∣ z + 1 ∣

\int \frac{- z + 2}{3 ( z ^{2} - z + 1 )} d z = \frac{1}{6} (- ln \frac{4}{3} + \frac{4 z ^{2} - 4 z}{3} + 23 arctan (\frac{2 z - 1}{3}))

= \frac{1}{3} ln ∣ z + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 z ^{2} - 4 z}{3} + 23 arctan (\frac{2 z - 1}{3}))

解答补常数

= \frac{1}{3} ln ∣ z + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 z ^{2} - 4 z}{3} + 23 arctan (\frac{2 z - 1}{3})) + C

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