求定积分：[1，+∞]∫arctanxdx/x²
解：令u=arctanx，则x=tanu，dx=du/cos²u，x=1时u=π/4；x→+∞时u=π/2；
故原式=[π/4，π/2]∫udu/(tan²ucos²u)=[π/4，π/2]∫udu/sin²u=[π/4，π/2][-∫udcotu]
=[π/4，π/2][-ucotu+∫cotudu]=[π/4，π/2][-ucotu+∫(cosu/sinu)du]
=[π/4，π/2][-ucotu+lnsinu]=(π/4)-ln(√2/2)=(π/4)+(1/2)ln2.
解：令u=arctanx，则x=tanu，dx=du/cos²u，x=1时u=π/4；x→+∞时u=π/2；
故原式=[π/4，π/2]∫udu/(tan²ucos²u)=[π/4，π/2]∫udu/sin²u=[π/4，π/2][-∫udcotu]
=[π/4，π/2][-ucotu+∫cotudu]=[π/4，π/2][-ucotu+∫(cosu/sinu)du]
=[π/4，π/2][-ucotu+lnsinu]=(π/4)-ln(√2/2)=(π/4)+(1/2)ln2.
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第1个回答 2011-12-02
let
tana = x
(seca)^2 da = dx
x= 无穷, a = π/2
x=1, a=π/4
∫(arctanx / x^2 ) dx
= ∫ a da a =(π/4,π/2)
= [a^2/2] a =(π/4,π/2)
= 3π^2/16
tana = x
(seca)^2 da = dx
x= 无穷, a = π/2
x=1, a=π/4
∫(arctanx / x^2 ) dx
= ∫ a da a =(π/4,π/2)
= [a^2/2] a =(π/4,π/2)
= 3π^2/16
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