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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

因式分解

u^{2} - 1 : - (- u^{2} + 1)

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

提出常数:

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

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

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

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

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

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

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

\int 1 d u = u

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

u = x

代回

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

化简

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

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

解答补常数

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

作图

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t \frac{d y}{d t} + 34 y = \frac{1}{t}

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