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受欢迎的微积分 >

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

解答

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

解答

- 2 (- 1 + x + ln ∣ 1 - x ∣) + C

求解步骤

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

乘开

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

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

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 u + \int \frac{1}{u} d u))

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (- (- u + ln ∣ u ∣))

u = 1 - x

代回

= 2 (- (- (1 - x) + ln ∣ 1 - x ∣))

化简

= - 2 (- 1 + x + ln ∣ 1 - x ∣)

解答补常数

= - 2 (- 1 + x + ln ∣ 1 - x ∣) + C

作图

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