计算定积分 ∫（1→根号3）[1/{x²根号下（1+x²）}]dx

此题用三角代换（换元法）
令x=tant, 则 dx=sec²tdt
∵x∈[1,√3]
∴不妨令t∈[π/4,π/3](在此区间上,x随t单增,sect≥0)
原积分=∫(π/4,π/3) sec²tdt/(tan²t·sect)
=∫(π/4,π/3) sectdt/tan²t
=∫(π/4,π/3) dt/(tan²t·cost)
=∫(π/4,π/3) costdt/sin²t
=∫(π/4,π/3) d(sint)/sin²t
=[-1/sint]|(π/4,π/3)
=√2-2/√3
=√2-√6/3
希望我的解答对你有所帮助√2-2/√3和√2-√6/3不相等的好伐。。但上边的都对哦谢谢