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受欢迎的微积分 >

integral of 1/(sqrt(1-2e^x))

解答

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

解答

- ln 1 - 2 e^{x} + 1 + ln 1 - 2 e^{x} - 1 + C

求解步骤

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

使用换元积分法

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

使用换元积分法

= \int \frac{2}{v ^{2} - 1} d v

提出常数:

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

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

因式分解

v^{2} - 1 : - (- v^{2} + 1)

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

提出常数:

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

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

使用积分常见定则:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= 2 (- (\frac{ln 1 - 2 e ^{x} + 1}{2} - \frac{ln 1 - 2 e ^{x} - 1}{2}))

化简

2 (- (\frac{ln 1 - 2 e ^{x} + 1}{2} - \frac{ln 1 - 2 e ^{x} - 1}{2})) : - ln 1 - 2 e^{x} + 1 + ln 1 - 2 e^{x} - 1

= - ln 1 - 2 e^{x} + 1 + ln 1 - 2 e^{x} - 1

解答补常数

= - ln 1 - 2 e^{x} + 1 + ln 1 - 2 e^{x} - 1 + C

作图

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