求不定积分∫1/(1+根号下1-x平方)
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答：
∫dx/[1+√(1-x^2)] 设x=sint，-π/2<=t<=π/2
=∫[1/(1+cost)]d(sint)
=∫[(1+cost-1)/(1+cost)]dt
=∫[1-1/(1+cost)]dt
=t-∫(1/cos²t/2)d(t/2)
=t+cot(t/2)+C
=arcsinx+[1+√(1-x^2)]/x+C
∫dx/[1+√(1-x^2)] 设x=sint，-π/2<=t<=π/2
=∫[1/(1+cost)]d(sint)
=∫[(1+cost-1)/(1+cost)]dt
=∫[1-1/(1+cost)]dt
=t-∫(1/cos²t/2)d(t/2)
=t+cot(t/2)+C
=arcsinx+[1+√(1-x^2)]/x+C
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