daifeifan_1026
幼苗
共回答了11个问题采纳率:90.9% 举报
∫[1→√3] 1/[x²√(1+x²)] dx
令x=tanu,则√(1+x²)=secu,dx=sec²udu,u:π/4→π/3
=∫[π/4→π/3] [1/(tan²usecu)](sec²u) du
=∫[π/4→π/3] secu/tan²u du
=∫[π/4→π/3] cosu/sin²u du
=∫[π/4→π/3] 1/sin²u dsinu
=-1/sinu ||[π/4→π/3]
=√2 - 2/√3
1年前
5