计算定积分, 要求有必要具体过程, 题目内容见图.

Question

如题所述

Answer 1

x = tanu，dx = sec²u du
x = 1，u = π/4
x = √3，u = π/3
[1, √3] ∫ dx / [x² √(1+x²)]
= [π/4, π/3] ∫ sec²u du / [tan²u √(1 + tan²u)]
= [π/4, π/3] ∫ cosu du / sin²u
= [π/4, π/3] ∫ dsinu / sin²u
= -1/sinu | [π/4, π/3]
=1/sin(π/4) - 1/sin(π/3)
=√2 - 2√3 /3