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受欢迎的微积分 >

integral of sqrt(10-e^{2x)}

解答

\int 10 - e^{2 x} d x

解答

- 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x})) + C

求解步骤

\int 10 - e^{2 x} d x

使用换元积分法

= \int \frac{10 - u}{2 u} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{10 - u}{u} d u

使用换元积分法

= \frac{1}{2} \cdot \int - \frac{2 v ^{2}}{- v ^{2} + 10} d v

提出常数:

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

= \frac{1}{2} (- 2 \cdot \int \frac{v ^{2}}{- v ^{2} + 10} d v)

使用换元积分法

= \frac{1}{2} (- 2 \cdot \int \frac{10 w ^{2}}{- w ^{2} + 1} d w)

提出常数:

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

= \frac{1}{2} (- 210 \cdot \int \frac{w ^{2}}{- w ^{2} + 1} d w)

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

= \frac{1}{2} (- 210 \cdot \int \frac{1}{- w ^{2} + 1} - 1 d w)

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

= \frac{1}{2} (- 210 (\int \frac{1}{- w ^{2} + 1} d w - \int 1 d w))

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

\int 1 d w = w

= \frac{1}{2} (- 210 (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2} - w))

代回

= \frac{1}{2} - 210 \frac{ln \frac{10 - e ^{2 x}}{10} + 1}{2} - \frac{ln \frac{10 - e ^{2 x}}{10} - 1}{2} - \frac{10 - e ^{2 x}}{10}

化简

\frac{1}{2} - 210 \frac{ln \frac{10 - e ^{2 x}}{10} + 1}{2} - \frac{ln \frac{10 - e ^{2 x}}{10} - 1}{2} - \frac{10 - e ^{2 x}}{10} : - 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x}))

= - 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x}))

解答补常数

= - 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x})) + C

作图

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