答：∫ {1/[x²√(1+x²)]}dx 设x=tant∈[1，√3]，π/4<=t<=π/3=∫{1/[tan²t*1/cost]}d(tant)=∫ (cos³t/sin²t)*(1/cos²t)dt=∫(cost/sin²t)dt=∫(1/sin²t)d(sint)=-1/sint+C所以定积分=-√2-(-2/√3)=2/√3-√2=2√3/3-√2所以定积分=2√3/3-√2
