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
