∫√(1-x^2) /x dx
=∫x√(1-x^2) /x² dx
=(1/2)∫√(1-x^2) /x² dx²
令√(1-x^2)=u，则1-x²=u²，dx²=-du²=-2udu
=(1/2)∫ -2u²/(1-u²) du
=∫ u²/(u²-1²) du
=∫ (u²-1+1)/(u²-1²) du
=∫ (1+1/(u²-1²)) du
=u + (1/2)ln|(u-1)/(u+1)| + C
=√(1-x²) + (1/2)ln|(√(1-x²)-1)/(√(1-x²)+1)| + C
令x=sint
∫√(1-x^2) /x dx=∫cos^2t/sintdt
=∫(1/sint-sint)dt
=ln|csc t-cot t|+cos t+C
=ln|1/x-√(1-x^2)/x|+√(1-x^2)+C
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