∫1/[x-√(1-x^2)]dx

令x=sint,dx=cost∫1/[x-√(1-x^2)]dx=∫cost/(sint-cost)dt令cost=a[sint-cost]+b[sint-cost]'=a[sint-cost]+b[cost+sint]=(a+b)sint+(b-a)cost则a+b=0,b-a=1,解得a=-1/2,b=1/2那么∫cost/(sint-cost)dt=(1/2)∫[-(...