这道定积分的题怎么做？

Question

如题所述

Answer 1

换元法求：令 (1 + x) / √(3x^2 - 2x - 1) = t

两边平方得到：(x^2 + 2x + 1) / (3x^2 - 2x - 1) = t^2
=> t^2 + 1 = 4 * x^2 / (3x^2 - 2x - 1)
两边同时微分并且化简得到：-4x / (3x^2 - 2x - 1)^(3/2) dx = dt
=>dx = - (3x^2 - 2x - 1)^(3/2) / (4x) dt
x=1, t=+∞， x=2, t=3/√7
则代入原积分 = - ∫ 1 /（x*（3x^2-2x-1）^1/2）* (3x^2 - 2x - 1)^(3/2) / (4x) dt
= - ∫ (3x^2 - 2x - 1) / (4 * x^2) dt
= - ∫ 1 / (1 + t^2) dt
= - arctan t |(+∞，3/√7)
=π/2-arctan(3/√7)