求定积分∫dx/[(x^2)*((1+x^2)^(1/2))],其中积分上限是根号3,积分下限是1,求详细过程~

问答/479℃/2025-05-26 11:28:19

优质解答：

F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))]

设x=tant,则dx=sec²tdt,∵x∈[1,√3],∴t∈[π/4,π/3]

∴F(x)=∫dx/[(x^2)*((1+x^2)^(1/2))] x∈[1,√3]

=∫sec²tdt/[tan²t*(1+tan²t)^(1/2)] t∈[π/4,π/3]

=∫sec²tdt/(tan²t*sect)

=∫sectdt/tan²t

=∫(cos²t/sin²t)*(1/cost)*dt

=∫(cost/sin²t)dt

=∫dsint/sin²t

=(-1/sint) t∈[π/4,π/3]

=1/sin(π/4)-1/sin(π/3)

=2/√2-2/√3

=(3√2-2√3)/3

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