高数定积分

Question

如题所述

Answer 1

解：
令t=arctanx，则x=tant
x：1→+∞，则t：π/4→π/2
∫[1：+∞](arctanx/x²)dx
=∫[π/4：π/2](t/tan²t)d(tant)
=∫[π/4：π/2][t/tan²t)sec²t]dt
=∫[π/4：π/2](t·csc²t)dt
=-∫[π/4：π/2]td[cot(t)]
=-t·cot(t)|[π/4：π/2] +∫[π/4：π/2]cot(t)dt
=-(0-π/4)+∫[π/4：π/2](cost/sint)dt
=¼Ï€ +ln|sint||[π/4：π/2]
=¼Ï€+ln|sin(π/2)|-ln|sin(π/4)|
=¼Ï€+ln1-ln|1/√2|
=¼Ï€+0+½ln2
=¼Ï€ +½ln2

Answer 2

∫arctanx/x²dx
=-∫arctanxd(1/x)
=-arctanx/x+∫1/x*1/(1+x²)dx
∫1/[x(1+x²)]dx
(令x=tanu)
=∫1/tanu*1/sec²u*sec²udu
=∫cosu/sinudu
=ln|sinu|+C
=ln|x/√(x²+1)|+C
所以
∫[1,+∞)arctanx/x²dx
=[-arctanx/x+ln(x/√(x²+1))] |[1,+∞)
=π/4-1/2*ln2追答

π/4+1/2*ln2