Question

高数 定积分 求大神给我检查一下这道题

Answer 1

积分区间忘了改
x=a tant∈(a，√3a)
tan t∈(1,√3)
t∈(π/4，π/3)

追问

谢谢你啊！一下就说明白了

Answer 2

let
x=atanu
dx=a(secu)^2 du

x=a, u=π/4
x=a√3, u =π/3

∫(a->a√3) dx/(a^2+x^2)
=(1/a)∫(π/4->π/3) du
=(1/a) [u] |(π/4->π/3)
= π/(12a)