integral of (x^3-x^2-4x+4)/(x^3-x^2-6x)

解答

\int \frac{x ^{3} - x ^{2} - 4 x + 4}{x ^{3} - x ^{2} - 6 x} d x

x - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣ + \frac{4}{5} ln ∣ x + 2 ∣ - \frac{17}{15} ln ∣ x - 3 ∣ - \frac{2}{3} ln ∣ x ∣ - \frac{1}{2} + C

求解步骤

\int \frac{x ^{3} - x ^{2} - 4 x + 4}{x ^{3} - x ^{2} - 6 x} d x

乘开

\frac{x ^{3} - x ^{2} - 4 x + 4}{x ^{3} - x ^{2} - 6 x} : \frac{x ^{3}}{x ^{3} - x ^{2} - 6 x} - \frac{x ^{2}}{x ^{3} - x ^{2} - 6 x} - \frac{4 x}{x ^{3} - x ^{2} - 6 x} + \frac{4}{x ^{3} - x ^{2} - 6 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{x ^{3}}{x ^{3} - x ^{2} - 6 x} d x - \int \frac{x ^{2}}{x ^{3} - x ^{2} - 6 x} d x - \int \frac{4 x}{x ^{3} - x ^{2} - 6 x} d x + \int \frac{4}{x ^{3} - x ^{2} - 6 x} d x

\int \frac{x ^{3}}{x ^{3} - x ^{2} - 6 x} d x = x - \frac{1}{2} - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣

\int \frac{x ^{2}}{x ^{3} - x ^{2} - 6 x} d x = \frac{2}{5} ln ∣ x + 2 ∣ + \frac{3}{5} ln ∣ x - 3 ∣

\int \frac{4 x}{x ^{3} - x ^{2} - 6 x} d x = 4 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣)

\int \frac{4}{x ^{3} - x ^{2} - 6 x} d x = 4 (- \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣)

= x - \frac{1}{2} - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣ - (\frac{2}{5} ln ∣ x + 2 ∣ + \frac{3}{5} ln ∣ x - 3 ∣) - 4 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣) + 4 (- \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣)

化简

x - \frac{1}{2} - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣ - (\frac{2}{5} ln ∣ x + 2 ∣ + \frac{3}{5} ln ∣ x - 3 ∣) - 4 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣) + 4 (- \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣) : x - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣ + \frac{4}{5} ln ∣ x + 2 ∣ - \frac{17}{15} ln ∣ x - 3 ∣ - \frac{2}{3} ln ∣ x ∣ - \frac{1}{2}

= x - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣ + \frac{4}{5} ln ∣ x + 2 ∣ - \frac{17}{15} ln ∣ x - 3 ∣ - \frac{2}{3} ln ∣ x ∣ - \frac{1}{2}

解答补常数

= x - \frac{4}{5} ln ∣ 2 x + 4 ∣ + \frac{9}{5} ln ∣ 2 x - 6 ∣ + \frac{4}{5} ln ∣ x + 2 ∣ - \frac{17}{15} ln ∣ x - 3 ∣ - \frac{2}{3} ln ∣ x ∣ - \frac{1}{2} + C

\frac{\partial}{\partial x} (e^{2 y} cos (3 x))

in v erse l a pl a ce 3

y^{^{''}} - 5 y^{^{'}} + 6 y = x \cdot e^{(- 4 \cdot x)}

\frac{d}{d x} (\frac{5}{( 2 x ) ^{3}})

\frac{\partial}{\partial x} (x^{2} - y^{2})