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

解答

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

\frac{1}{2 x ^{2}} + \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)) + C

求解步骤

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

将

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

用部份分式展开:

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

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

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

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

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

化简

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

解答补常数

= \frac{1}{2 x ^{2}} + \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)) + C

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}

\int (x^{2} - 9 x + 3) d x

\frac{\partial}{\partial x} (x y + ln (\frac{x}{y}))

\int \frac{- 16 x + 3}{( x ^{2} + 2 ) ^{2}} d x

\frac{\partial}{\partial x} (e^{(3 x + 4 y)} cos (5 z))