∫dx/[(1+x^2)^2]
令x=tanα α∈(-π/2，π/2)
cosα=[1/(1+x^2)]^(1/2)
sinα=x[1/(1+x^2)]^(1/2)
∫dx/[(1+x^2)^2]
=∫dtanα/[(secα)^4]
=∫dα/(secα)^2
=∫(cosα)^2dα
=0.5∫(cos2α + 1)dα
=0.5∫cos2αdα + 0.5∫dα
=(sin2α)/4 + α/2 + C
=sinαcosα/2 + α/2 + C
=x/[2(1+x^2)] + 1/2 arctanx + C
1/[(1+x^2)^2]在[0，+∞]上的定积分
=lim(x→+∞) x/[2(1+x^2)] + 1/2 lim(x→+∞) arctanx
=0+1/2 × π/2
=π/4
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