令 x = tant, dx = (sect)^2dt
当 x→0，t→0；当 x→+∞，t→π/2
∫[0,+∞]dx/(1+x^2)^2
=∫[0,π/2](sect)^2dt/(sect)^4
=∫[0,π/2]dt/(sect)^2
=∫[0,π/2](cost)^2dt
=1/2∫[0,π/2](1+cos2t)dt
=1/2(t+1/2sin2t)[0,π/2]
=π/4
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令 x = tant, dx = (sect)^2dt
当 x→0，t→0；当 x→+∞，t→π/2
∫[0,+∞]dx/(1+x^2)^2
=∫[0,π/2](sect)^2dt/(sect)^4
=∫[0,π/2]dt/(sect)^2
=∫[0,π/2](cost)^2dt
=1/2∫[0,π/2](1+cos2t)dt
=1/2(t+1/2sin2t)[0,π/2]
=π/4
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