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∫arctanxdx/x^2上限是无穷,下限是1.怎么求啊
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∫arctanxdx/x^2 上限是无穷,下限是1.怎么求啊
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求定积分：[1,+∞]∫arctanxdx/x²
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.
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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