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受欢迎的微积分 >

integral of 9sqrt(1+e^x)

解答

\int 9 1 + e^{x} d x

解答

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

求解步骤

\int 9 1 + e^{x} d x

提出常数:

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

= 9 \cdot \int 1 + e^{x} d x

使用换元积分法

= 9 \cdot \int \frac{1 + u}{u} d u

使用换元积分法

= 9 \cdot \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

= 9 \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} - 1} d v

因式分解

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

= 9 \cdot 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

= 9 \cdot 2 (- \int \frac{v ^{2}}{- v ^{2} + 1} d v)

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

= 9 \cdot 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

= 9 \cdot 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

= 9 \cdot 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

代回

= 9 \cdot 2 (- (\frac{ln 1 + e ^{x} + 1}{2} - \frac{ln 1 + e ^{x} - 1}{2} - 1 + e^{x}))

化简

9 \cdot 2 (- (\frac{ln 1 + e ^{x} + 1}{2} - \frac{ln 1 + e ^{x} - 1}{2} - 1 + e^{x})) : - 18 (\frac{1}{2} ln 1 + e^{x} + 1 - \frac{1}{2} ln 1 + e^{x} - 1 - 1 + e^{x})

= - 18 (\frac{1}{2} ln 1 + e^{x} + 1 - \frac{1}{2} ln 1 + e^{x} - 1 - 1 + e^{x})

解答补常数

= - 18 (\frac{1}{2} ln 1 + e^{x} + 1 - \frac{1}{2} ln 1 + e^{x} - 1 - 1 + e^{x}) + C

作图

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