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设arctanx=t，则x=tant.
∫(0,1)arctanxdx/(1+x^2)^2
=∫(0,π/4)tdtant/(1+tan^2t)^2
=∫(0,π/4)tsec^2tdt/sec^4t
=∫(0,π/4)tdt/sec^2t
=∫(0,π/4)tcos^2tdt
=∫(0,π/4)t(1+cos2t)dt/2
=(1/2)∫(0,π/4)tdt+(1/4)∫(0,π/4)tcos2td2t
=(1/4)t^2(0,π/4)+(1/4)tsin2t(0,π/4)-(1/4)∫(0,π/4)sin2tdt
=π^2/64+π/16+(1/8)cos2t(0,π/4)
=π^2/64+π/16-(1/8)