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受欢迎的微积分 >

integral of ((4-sqrt(2x+1)))/(1-2x)

解答

\int \frac{( 4 - 2 x + 1 )}{1 - 2 x} d x

解答

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

求解步骤

\int \frac{4 - 2 x + 1}{1 - 2 x} d x

使用换元积分法

= \int \frac{4 - u}{2 ( - u + 2 )} 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{4 - u}{- u + 2} d u

乘开

\frac{4 - u}{- u + 2} : \frac{4}{- u + 2} - \frac{u}{- u + 2}

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} (\int \frac{4}{- u + 2} d u - \int \frac{u}{- u + 2} d u)

\int \frac{4}{- u + 2} d u = - 4 ln ∣ - u + 2 ∣

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

= \frac{1}{2} (- 4 ln ∣ - u + 2 ∣ - 2 (2 (\frac{1}{2} ln \frac{u}{2} + 1 - \frac{1}{2} ln \frac{u}{2} - 1) - u))

u = 2 x + 1

代回

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

化简

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

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

解答补常数

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

作图

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