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求助大神,这道题怎么做
1个回答
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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)
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)
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