大学高数题 200
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令arctanx=u，则x=tanu
x：1→∞，则u：π/4→π/2
∫[1：+∞] (arctanx/x²)dx
=∫[π/4：π/2] (u/tan²u)d(tanu)
=∫[π/4：π/2] (u/tan²u)·sec²udu
=∫[π/4：π/2] (u·csc²u)du
=-∫[π/4：π/2] ud(cotu)
=-ucotu|[π/4：π/2]+∫[π/4：π/2]cotudu
=-[(π/2)·cot(π/2)-(π/4)·cot(π/4)]+∫[π/4：π/2](cosu/sinu)du
=-[(π/2)·0-(π/4)·1]+∫[π/4：π/2](1/sinu)d(sinu)
=π/4 +ln|sinu||[π/4：π/2]
=π/4+ln|sinπ/2|-ln|sinπ/4|
=π/4+ln1 -ln(√2/2)
=π/4 +0 +½ln2
=(π+2ln2)/4
x：1→∞，则u：π/4→π/2
∫[1：+∞] (arctanx/x²)dx
=∫[π/4：π/2] (u/tan²u)d(tanu)
=∫[π/4：π/2] (u/tan²u)·sec²udu
=∫[π/4：π/2] (u·csc²u)du
=-∫[π/4：π/2] ud(cotu)
=-ucotu|[π/4：π/2]+∫[π/4：π/2]cotudu
=-[(π/2)·cot(π/2)-(π/4)·cot(π/4)]+∫[π/4：π/2](cosu/sinu)du
=-[(π/2)·0-(π/4)·1]+∫[π/4：π/2](1/sinu)d(sinu)
=π/4 +ln|sinu||[π/4：π/2]
=π/4+ln|sinπ/2|-ln|sinπ/4|
=π/4+ln1 -ln(√2/2)
=π/4 +0 +½ln2
=(π+2ln2)/4
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