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~

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

求定积分 1/x^2(1+x^2)=1/x^2-1/（1+x^2) 用公式求出1/x^2和1/（1+x^2)定积分然后将上下限代进去即可

原式=∫<π/4,π/3>sec²tdt/(tan²t*sect) (令x=tant)
=∫<π/4,π/3>costdt/sin²t
=∫<π/4,π/3>d(sint)/sin²t
=(-1/sint)│<π/4,π/3>
=√2-2√3/3。