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

解答

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

\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 ^{3} - 1} d x

将

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

用部份分式展开:

\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

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

= \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}{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}{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}{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}{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

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

\frac{d}{d x} (tan (1 - x))

\frac{\partial}{\partial x} (x^{2} + 10 x y + y^{2})

x \to 0 + lim (x \cdot (e^{\frac{1}{x}} - 1))

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