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受欢迎的微积分 >

integral of 1/(sqrt(4-e^{2x))}

解答

\int \frac{1}{4 - e ^{2 x}} d x

解答

- \frac{1}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1) + C

求解步骤

\int \frac{1}{4 - e ^{2 x}} d x

使用换元积分法

= \int - \frac{1}{- u ^{2} + 4} d u

提出常数:

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

= - \int \frac{1}{- u ^{2} + 4} d u

使用换元积分法

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

提出常数:

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

= - \frac{1}{2} \cdot \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}

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

代回

= - \frac{1}{2} \frac{ln \frac{4 - e ^{2 x}}{2} + 1}{2} - \frac{ln \frac{4 - e ^{2 x}}{2} - 1}{2}

化简

- \frac{1}{2} \frac{ln \frac{4 - e ^{2 x}}{2} + 1}{2} - \frac{ln \frac{4 - e ^{2 x}}{2} - 1}{2} : - \frac{1}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1)

= - \frac{1}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1)

解答补常数

= - \frac{1}{4} (ln \frac{4 - e ^{2 x}}{2} + 1 - ln \frac{4 - e ^{2 x}}{2} - 1) + C

作图

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