integral of (6x+2)/((x+1)(2x+1))

解答

\int \frac{6 x + 2}{( x + 1 ) ( 2 x + 1 )} d x

6 ln ∣ 2 x + 2 ∣ - ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣ + C

求解步骤

\int \frac{6 x + 2}{( x + 1 ) ( 2 x + 1 )} d x

乘开

\frac{6 x + 2}{( x + 1 ) ( 2 x + 1 )} : \frac{6 x}{( x + 1 ) ( 2 x + 1 )} + \frac{2}{( x + 1 ) ( 2 x + 1 )}

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

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

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

= 3 (2 ln ∣ 2 x + 2 ∣ - ln ∣ 2 x + 1 ∣) + 2 (- ln ∣ x + 1 ∣ + ln ∣ 2 x + 1 ∣)

化简

3 (2 ln ∣ 2 x + 2 ∣ - ln ∣ 2 x + 1 ∣) + 2 (- ln ∣ x + 1 ∣ + ln ∣ 2 x + 1 ∣) : 6 ln ∣ 2 x + 2 ∣ - ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣

= 6 ln ∣ 2 x + 2 ∣ - ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣

解答补常数

= 6 ln ∣ 2 x + 2 ∣ - ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣ + C

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