integral of 1/(x^6-8x^3)

解答

\int \frac{1}{x ^{6} - 8 x ^{3}} d x

\frac{1}{16 x ^{2}} + \frac{1}{96} ln ∣ x - 2 ∣ - \frac{1}{192} ln x^{2} + 2 x + 4 - \frac{1}{32 3} arctan (\frac{1}{3} (x + 1)) + C

求解步骤

\int \frac{1}{x ^{6} - 8 x ^{3}} d x

将

\frac{1}{x ^{6} - 8 x ^{3}}

用部份分式展开:

- \frac{1}{8 x ^{3}} + \frac{1}{96 ( x - 2 )} + \frac{- x - 4}{96 ( x ^{2} + 2 x + 4 )}

= \int - \frac{1}{8 x ^{3}} + \frac{1}{96 ( x - 2 )} + \frac{- x - 4}{96 ( x ^{2} + 2 x + 4 )} d x

使用积分加法定则:

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

= - \int \frac{1}{8 x ^{3}} d x + \int \frac{1}{96 ( x - 2 )} d x + \int \frac{- x - 4}{96 ( x ^{2} + 2 x + 4 )} d x

\int \frac{1}{8 x ^{3}} d x = - \frac{1}{16 x ^{2}}

\int \frac{1}{96 ( x - 2 )} d x = \frac{1}{96} ln ∣ x - 2 ∣

\int \frac{- x - 4}{96 ( x ^{2} + 2 x + 4 )} d x = \frac{1}{192 3} (- 3 ln x^{2} + 2 x + 4 - 6 arctan (\frac{1}{3} (x + 1)))

= - (- \frac{1}{16 x ^{2}}) + \frac{1}{96} ln ∣ x - 2 ∣ + \frac{1}{192 3} (- 3 ln x^{2} + 2 x + 4 - 6 arctan (\frac{1}{3} (x + 1)))

化简

- (- \frac{1}{16 x ^{2}}) + \frac{1}{96} ln ∣ x - 2 ∣ + \frac{1}{192 3} (- 3 ln x^{2} + 2 x + 4 - 6 arctan (\frac{1}{3} (x + 1))) : \frac{1}{16 x ^{2}} + \frac{1}{96} ln ∣ x - 2 ∣ - \frac{1}{192} ln x^{2} + 2 x + 4 - \frac{1}{32 3} arctan (\frac{1}{3} (x + 1))

= \frac{1}{16 x ^{2}} + \frac{1}{96} ln ∣ x - 2 ∣ - \frac{1}{192} ln x^{2} + 2 x + 4 - \frac{1}{32 3} arctan (\frac{1}{3} (x + 1))

解答补常数

= \frac{1}{16 x ^{2}} + \frac{1}{96} ln ∣ x - 2 ∣ - \frac{1}{192} ln x^{2} + 2 x + 4 - \frac{1}{32 3} arctan (\frac{1}{3} (x + 1)) + C

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