integral of 1/(1-x^3)

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

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