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

解答

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

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

求解步骤

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

使用换元积分法

= \int - \frac{- u + 2}{u} d u

提出常数:

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

= - \int \frac{- u + 2}{u} d u

使用分布积分法

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

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

= - (\frac{- u + 2}{u} u - (- 2 arcsin (\frac{1}{2} u)))

u = 1 - x

代回

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

化简

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

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

解答补常数

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

\int sec^{4} (x) \cdot tan^{3} (x) d x

\int t^{\frac{3}{2}} (t^{2} - 10 t + 16) d t

\int \frac{( ln ^{7} ( x ) )}{x} d x

\int ln (2 x - 5) d x

\int e^{8 x} sin^{2} (x) d x