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解:∫(1,+∞) arctanx/x^3 dx
=-1/2*∫(1,+∞) arctanxd(1/锋首肢芹老x^2)
=-1/2*(arctanx)/x^2 |(1,+∞)+1/2*∫(1,+∞) (1/x^2)*1/(1+x^2)dx
=π/8-1/2*lim x->+∞ (arctanx)/x^2 +1/2*[∫(1,+∞)1/x^2dx-∫(1,+∞)1/(1+x^2)dx]
=π/银世8-1/2*lim x->+∞ 1/(1+x^2)/(2x) +1/2*[-1/x |(1,+∞)-arctanx|(1,+∞)]
=π/8-0 +1/2*{[0-(-1)]-lim x->+∞ arctanx+arctan1}
=π/8+1/2*(1-π/2+π/4)
=1/2
不明白请追问。
=-1/2*∫(1,+∞) arctanxd(1/锋首肢芹老x^2)
=-1/2*(arctanx)/x^2 |(1,+∞)+1/2*∫(1,+∞) (1/x^2)*1/(1+x^2)dx
=π/8-1/2*lim x->+∞ (arctanx)/x^2 +1/2*[∫(1,+∞)1/x^2dx-∫(1,+∞)1/(1+x^2)dx]
=π/银世8-1/2*lim x->+∞ 1/(1+x^2)/(2x) +1/2*[-1/x |(1,+∞)-arctanx|(1,+∞)]
=π/8-0 +1/2*{[0-(-1)]-lim x->+∞ arctanx+arctan1}
=π/8+1/2*(1-π/2+π/4)
=1/2
不明白请追问。
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