integral of 1/(s^3-1)

解

\int \frac{1}{s ^{3} - 1} d s

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

解答ステップ

\int \frac{1}{s ^{3} - 1} d s

以下の部分分数を得る:

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

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{3 ( s - 1 )} d s = \frac{1}{3} ln ∣ s - 1 ∣

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

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

簡素化

\frac{1}{3} ln ∣ s - 1 ∣ + \frac{1}{6 3} (- 3 ln 4 s^{2} + 4 s + 4 - 6 arctan (\frac{2 s + 1}{3})) : \frac{1}{3} ln ∣ s - 1 ∣ - \frac{1}{6} ln 4 s^{2} + 4 s + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 s + 1))

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

定数を解答に追加する

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