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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用换元积分法

= \int \frac{1}{ln ( 2 ) u ( u - 1 )} d u

提出常数:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{u ( u - 1 )} d u

使用换元积分法

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

提出常数:

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

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

将

\frac{1}{v ( v - 1 )}

用部份分式展开:

- \frac{1}{v} + \frac{1}{v - 1}

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

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

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

代回

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

化简

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

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

解答补常数

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

作图

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