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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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