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受欢迎的微积分 >

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

解答

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

解答

2 x - 1 + ln 2 x - 1 + C

求解步骤

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

使用换元积分法

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

使用换元积分法

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

提出常数:

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

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

乘开

\frac{v + 1}{v} : 1 + \frac{1}{v}

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{2} \cdot 2 (v + ln ∣ v ∣)

代回

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

化简

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

= 2 x - 1 + ln 2 x - 1

解答补常数

= 2 x - 1 + ln 2 x - 1 + C

作图

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