∫1/1-x^2dx=

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∫1/1-x^2dx=

∫1/1-x^2dx=
∫1/1-x^2dx=

∫1/1-x^2dx=
答:
∫[1/(1-x^2)]dx
=∫[1/(1-x)(1+x)]dx
=(1/2) ∫ [1/(1-x)+1/(1+x)]dx
=(1/2) ∫ [1/(1+x)] d(1+x) - (1/2) ∫ [1/(1-x)]d(1-x)
=(1/2)ln(1+x)-(1/2)ln(1-x)+C
=(1/2)ln[(1+x)/(1-x)]+C