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受欢迎的微积分 >

integral of sqrt(e^x+1)

解答

\int e^{x} + 1 d x

解答

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

求解步骤

\int e^{x} + 1 d x

使用换元积分法

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

使用换元积分法

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

因式分解

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

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

\int 1 d v = v

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

代回

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

化简

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

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

解答补常数

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

作图

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