integral of (x-5)/(x^3-3x^2+7x-5)

解答

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

- ln ∣ x - 1 ∣ - \frac{1}{8} ln \frac{1}{4} (x - 1)^{2} + 1 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + \frac{5}{8} ln x^{2} - 2 x + 5 + C

求解步骤

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

乘开

\frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} : \frac{x}{x ^{3} - 3 x ^{2} + 7 x - 5} - \frac{5}{x ^{3} - 3 x ^{2} + 7 x - 5}

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}{x ^{3} - 3 x ^{2} + 7 x - 5} d x - \int \frac{5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

\int \frac{x}{x ^{3} - 3 x ^{2} + 7 x - 5} d x = \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln \frac{1}{4} (x - 1)^{2} + 1 + 2 arctan (\frac{1}{2} (x - 1)))

\int \frac{5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x = 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{8} ln x^{2} - 2 x + 5)

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln \frac{1}{4} (x - 1)^{2} + 1 + 2 arctan (\frac{1}{2} (x - 1))) - 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{8} ln x^{2} - 2 x + 5)

化简

\frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln \frac{1}{4} (x - 1)^{2} + 1 + 2 arctan (\frac{1}{2} (x - 1))) - 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{8} ln x^{2} - 2 x + 5) : - ln ∣ x - 1 ∣ - \frac{1}{8} ln \frac{1}{4} (x - 1)^{2} + 1 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + \frac{5}{8} ln x^{2} - 2 x + 5

= - ln ∣ x - 1 ∣ - \frac{1}{8} ln \frac{1}{4} (x - 1)^{2} + 1 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + \frac{5}{8} ln x^{2} - 2 x + 5

解答补常数

= - ln ∣ x - 1 ∣ - \frac{1}{8} ln \frac{1}{4} (x - 1)^{2} + 1 + \frac{1}{2} arctan (\frac{1}{2} (x - 1)) + \frac{5}{8} ln x^{2} - 2 x + 5 + C

\frac{d}{d x} (arcsin (9 x) + arccos (9 x))

\int \frac{x ^{3}}{a} - \frac{a}{x ^{3}} d x

\frac{d}{d x} (x^{22})

2 x y + (2 y + x^{2}) \cdot (\frac{d y}{d x}) = 0

\frac{d}{d x} (- e^{x^{2} - 5 x})