∫ 根号(1-x) ⼀ x 求这题的不定积分解题过程，谢谢了，还有这个是定积分里面的哪一种题型啊？

令z = √(1 - x)，x = 1 - z²，dx = - 2zdz
∫ √(1 - x)/x dx
= ∫ z/(1 - z²) * -2z dz
= 2∫ z²/(z² - 1) dz
= 2∫ (z² - 1 + 1)/(z² - 1) dz
= 2∫ dz + (2)(1/2)∫ [(z + 1) - (z - 1)]/(z² - 1) dz
= 2∫ dz + ∫[1/(z - 1) - 1/(z + 1)] dz
= 2z + ln| (z - 1)/(z + 1) | + C
= 2√(1 - x) + ln| [√(1 - x) - 1]/[√(1 - x) + 1] | + C

∫√(1-x)dx/x
x>0
x=sinu^2
dx=2sinucosudu
=∫2sinucosu^2du/sinu^2
=∫2cosu^2du/sinu
=∫(2-2sinu^2)du/sinu
=2∫du/sinu+2cosu
= -2∫dcosu/(1-cosu)(1+cosu) +2cosu+C
=-2ln|1+cosu|/|1-cosu| +2cosu+C
=-2ln|cscu+cotu|+2cosu+C
=-2ln|1/√x+√(1/x-1) +2√(1-x)+C

x<0
x=-sinu^2 dx=-2sinucosudu
=-2ln|cscu+cotu|+2cosu+C
=-2ln|1/√(-x)+√(1-x)/√(-x)|+2√(1-x)+C