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受欢迎的微积分 >

integral of (2u)/(1-u)

解答

\int \frac{2 u}{1 - u} d u

解答

- 2 (- 1 + u + ln ∣ 1 - u ∣) + C

求解步骤

\int \frac{2 u}{1 - u} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{u}{1 - u} d u

使用换元积分法

= 2 \cdot \int - \frac{- v + 1}{v} d v

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- \int \frac{- v + 1}{v} d v)

乘开

\frac{- v + 1}{v} : - 1 + \frac{1}{v}

= 2 (- \int - 1 + \frac{1}{v} d v)

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

= 2 (- (- \int 1 d v + \int \frac{1}{v} d v))

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (- (- v + ln ∣ v ∣))

v = 1 - u

代回

= 2 (- (- (1 - u) + ln ∣ 1 - u ∣))

化简

= - 2 (- 1 + u + ln ∣ 1 - u ∣)

解答补常数

= - 2 (- 1 + u + ln ∣ 1 - u ∣) + C

作图

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