∫1/[x-√(1-x^2)]dx

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∫1/[x-√(1-x^2)]dx∫1/[x-√(1-x^2)]dx∫1/[x-√(1-x^2)]dx令x=sint,dx=cost∫1/[x-√(1-x^2)]dx=∫cost/(sint-cost

∫1/[x-√(1-x^2)]dx
∫1/[x-√(1-x^2)]dx

∫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)∫[-(sint-cost)+(sint-cost)']/(sint-cost)dt
=-(1/2)∫dt+(1/2)∫1/(sint-cost)d(sint-cost)
=1/2[-t+ln|sint-cost|+C
三角代换t=arcsinx,sint=x,cost=√(1-x²)
所以
∫1/[x-√(1-x^2)]dx
=1/2[-arcsinx+ln|x-√(1-x²)|]+C

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