integral of 1/(x^3-512)

解

\int \frac{1}{x ^{3} - 512} d x

\frac{1}{192} ln ∣ x - 8 ∣ - \frac{1}{384} ln x^{2} + 8 x + 64 - \frac{1}{64 3} arctan (\frac{1}{4 3} x + \frac{1}{3}) + C

解答ステップ

\int \frac{1}{x ^{3} - 512} d x

以下の部分分数を得る:

\frac{1}{x ^{3} - 512} : \frac{1}{192 ( x - 8 )} + \frac{- x - 16}{192 ( x ^{2} + 8 x + 64 )}

= \int \frac{1}{192 ( x - 8 )} + \frac{- x - 16}{192 ( x ^{2} + 8 x + 64 )} 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}{192 ( x - 8 )} d x + \int \frac{- x - 16}{192 ( x ^{2} + 8 x + 64 )} d x

\int \frac{1}{192 ( x - 8 )} d x = \frac{1}{192} ln ∣ x - 8 ∣

\int \frac{- x - 16}{192 ( x ^{2} + 8 x + 64 )} d x = \frac{1}{384 3} (- 3 ln x^{2} + 8 x + 64 - 6 arctan (\frac{1}{4 3} x + \frac{1}{3}))

= \frac{1}{192} ln ∣ x - 8 ∣ + \frac{1}{384 3} (- 3 ln x^{2} + 8 x + 64 - 6 arctan (\frac{1}{4 3} x + \frac{1}{3}))

拡張

\frac{1}{384 3} (- 3 ln x^{2} + 8 x + 64 - 6 arctan (\frac{1}{4 3} x + \frac{1}{3})) : - \frac{1}{384} ln x^{2} + 8 x + 64 - \frac{1}{64 3} arctan (\frac{1}{4 3} x + \frac{1}{3})

= \frac{1}{192} ln ∣ x - 8 ∣ - \frac{1}{384} ln x^{2} + 8 x + 64 - \frac{1}{64 3} arctan (\frac{1}{4 3} x + \frac{1}{3})

定数を解答に追加する

= \frac{1}{192} ln ∣ x - 8 ∣ - \frac{1}{384} ln x^{2} + 8 x + 64 - \frac{1}{64 3} arctan (\frac{1}{4 3} x + \frac{1}{3}) + C