integral of 1/(x^3-64)

解答

\int \frac{1}{x ^{3} - 64} d x

\frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3}) + C

求解步骤

\int \frac{1}{x ^{3} - 64} d x

将

\frac{1}{x ^{3} - 64}

用部份分式展开:

\frac{1}{48 ( x - 4 )} + \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )}

= \int \frac{1}{48 ( x - 4 )} + \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} 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}{48 ( x - 4 )} d x + \int \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} d x

\int \frac{1}{48 ( x - 4 )} d x = \frac{1}{48} ln ∣ x - 4 ∣

\int \frac{- x - 8}{48 ( x ^{2} + 4 x + 16 )} d x = \frac{1}{96 3} (- 3 ln \frac{( x + 2 ) ^{2}}{12} + 1 - 6 arctan (\frac{1}{2 3} x + \frac{1}{3}))

= \frac{1}{48} ln ∣ x - 4 ∣ + \frac{1}{96 3} (- 3 ln \frac{( x + 2 ) ^{2}}{12} + 1 - 6 arctan (\frac{1}{2 3} x + \frac{1}{3}))

化简

\frac{1}{48} ln ∣ x - 4 ∣ + \frac{1}{96 3} (- 3 ln \frac{( x + 2 ) ^{2}}{12} + 1 - 6 arctan (\frac{1}{2 3} x + \frac{1}{3})) : \frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3})

= \frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3})

解答补常数

= \frac{1}{48} ln ∣ x - 4 ∣ - \frac{1}{96} ln \frac{1}{12} (x + 2)^{2} + 1 - \frac{1}{16 3} arctan (\frac{1}{2 3} x + \frac{1}{3}) + C

(\frac{π}{x})^{^{'}}

\frac{\partial}{\partial h} (\frac{π}{3} h^{2} (3 r - h))

\frac{d}{d x} (cos (x - 7 x^{2}))

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

\int \frac{tan ( x )}{cos ( x )} d x