integral of (2x+1)/((x-1)(x+2)(x+5))

解答

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

\frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 5 ∣ + C

求解步骤

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

乘开

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

\int \frac{2 x}{( x - 1 ) ( x + 2 ) ( x + 5 )} d x = 2 (\frac{1}{18} ln ∣ x - 1 ∣ + \frac{2}{9} ln ∣ x + 2 ∣ - \frac{5}{18} ln ∣ x + 5 ∣)

\int \frac{1}{( x - 1 ) ( x + 2 ) ( x + 5 )} d x = \frac{1}{18} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 2 ∣ + \frac{1}{18} ln ∣ x + 5 ∣

= 2 (\frac{1}{18} ln ∣ x - 1 ∣ + \frac{2}{9} ln ∣ x + 2 ∣ - \frac{5}{18} ln ∣ x + 5 ∣) + \frac{1}{18} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 2 ∣ + \frac{1}{18} ln ∣ x + 5 ∣

化简

2 (\frac{1}{18} ln ∣ x - 1 ∣ + \frac{2}{9} ln ∣ x + 2 ∣ - \frac{5}{18} ln ∣ x + 5 ∣) + \frac{1}{18} ln ∣ x - 1 ∣ - \frac{1}{9} ln ∣ x + 2 ∣ + \frac{1}{18} ln ∣ x + 5 ∣ : \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 5 ∣

= \frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 5 ∣

解答补常数

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

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