integral of (x^3+1)/(x^3-8)

解

\int \frac{x ^{3} + 1}{x ^{3} - 8} d x

x + \frac{3}{4} ln ∣ x - 2 ∣ - \frac{3}{8} ln x^{2} + 2 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (x + 1)) - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)) + C

解答ステップ

\int \frac{x ^{3} + 1}{x ^{3} - 8} d x

拡張

\frac{x ^{3} + 1}{x ^{3} - 8} : \frac{x ^{3}}{x ^{3} - 8} + \frac{1}{x ^{3} - 8}

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{x ^{3}}{x ^{3} - 8} d x + \int \frac{1}{x ^{3} - 8} d x

\int \frac{x ^{3}}{x ^{3} - 8} d x = x + \frac{2}{3} ln ∣ x - 2 ∣ + \frac{1}{3} (- ln x^{2} + 2 x + 4 - 23 arctan (\frac{x + 1}{3}))

\int \frac{1}{x ^{3} - 8} d x = \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))

= x + \frac{2}{3} ln ∣ x - 2 ∣ + \frac{1}{3} (- ln x^{2} + 2 x + 4 - 23 arctan (\frac{x + 1}{3})) + \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))

簡素化

x + \frac{2}{3} ln ∣ x - 2 ∣ + \frac{1}{3} (- ln x^{2} + 2 x + 4 - 23 arctan (\frac{x + 1}{3})) + \frac{1}{12} ln ∣ x - 2 ∣ - \frac{1}{24} ln x^{2} + 2 x + 4 - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)) : x + \frac{3}{4} ln ∣ x - 2 ∣ - \frac{3}{8} ln x^{2} + 2 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (x + 1)) - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))

= x + \frac{3}{4} ln ∣ x - 2 ∣ - \frac{3}{8} ln x^{2} + 2 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (x + 1)) - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1))

定数を解答に追加する

= x + \frac{3}{4} ln ∣ x - 2 ∣ - \frac{3}{8} ln x^{2} + 2 x + 4 - \frac{2}{3} arctan (\frac{1}{3} (x + 1)) - \frac{1}{4 3} arctan (\frac{1}{3} (x + 1)) + C