∫[上√3,下1]dx/x^2√1+x^2 的定积分详细过程

2025-01-21 02:01:02
推荐回答(2个)
回答1:

=(令x=tant,x1=√3,t=兀/3,X2=1,t=兀/4)∫(兀/4→兀/3)[1/(tant²√(1+tant²)]sect²dt
=∫(兀/4→兀/3)sect/tan²tdt
=∫(兀/4→兀/3)cost/sin²tdt
=∫(兀/4→兀/3)1/sin²td(sint)
=[-1/sint](兀/4→兀/3)
=-(2/√3一2/√2)
=√2-2√2/3

回答2:

∫(1→2) (x² + 1/x⁴) dx
= ∫(1→2) [x² + x^(-4)] dx
= [x³/3 + x^(-3)/(-3)] |(1→2)
= (1/3)[x³ - 1/x³] |(1→2)
= (1/3)[(8 - 1/8) - (1 - 1)]
= (1/3)(63/8)
= 63/24,63和24约分,各除以3
= 21/8