Question

求助大神，这道题怎么做

Answer 1

解：
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·sec²u/tan²u)du
=∫[π/4：π/2](u·csc²u)du
=-∫[π/4：π/2]ud(cotu)
=-u·cotu|[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
=¼(π+2ln2)