求定积分∫[1/x√(x²-1)]dx x属于（-2,-1）

网友回答

∫ 1/[x√(x² - 1)] dx,x∈[-2,-1]√(sec²θ - 1) = √(tan²θ) = ±tanθ令x = secθ,dx = secθtanθ dθ,θ∈[2π/3,π]∫ (secθtanθ)/(- secθtanθ) dθ= - θ= - [(π) - (2π/3)]= - π/3...