integral of (2x-1)/(x^3-6x^2+11x-6)

解答

\int \frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

\frac{1}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + C

求解步骤

\int \frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

将

\frac{2 x - 1}{x ^{3} - 6 x ^{2} + 11 x - 6}

用部份分式展开:

\frac{1}{2 ( x - 1 )} - \frac{3}{x - 2} + \frac{5}{2 ( x - 3 )}

= \int \frac{1}{2 ( x - 1 )} - \frac{3}{x - 2} + \frac{5}{2 ( x - 3 )} 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}{2 ( x - 1 )} d x - \int \frac{3}{x - 2} d x + \int \frac{5}{2 ( x - 3 )} d x

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

\int \frac{3}{x - 2} d x = 3 ln ∣ x - 2 ∣

\int \frac{5}{2 ( x - 3 )} d x = \frac{5}{2} ln ∣ x - 3 ∣

= \frac{1}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣

解答补常数

= \frac{1}{2} ln ∣ x - 1 ∣ - 3 ln ∣ x - 2 ∣ + \frac{5}{2} ln ∣ x - 3 ∣ + C

x \to - \infty lim (- \frac{6}{x})

y^{^{'}} + 2 y = e^{(4 t)}

\int (x^{2} - x) d x

\frac{d}{d x} (- 1.8 x^{2} + 14.4 x)

\frac{\partial}{\partial x} (- 4 (x + y^{6}))