integral of sqrt(4/x-1)

解答

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

x 4 - x - 4 arctan (\frac{4 - x}{x}) + C

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用分布积分法

= - 4 (- \frac{u - 1}{u} - \int - \frac{1}{2 u u - 1} d u)

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

= - 4 (- \frac{u - 1}{u} - (- arctan (u - 1)))

u = \frac{4}{x}

代回

= - 4 - \frac{\frac{4}{x} - 1}{\frac{4}{x}} - (- arctan (\frac{4}{x} - 1))

化简

- 4 - \frac{\frac{4}{x} - 1}{\frac{4}{x}} - (- arctan (\frac{4}{x} - 1)) : x 4 - x - 4 arctan (\frac{4 - x}{x})

= x 4 - x - 4 arctan (\frac{4 - x}{x})

解答补常数

= x 4 - x - 4 arctan (\frac{4 - x}{x}) + C