求反常积分
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换元x=√(u-1)，
=∫(1到∞)1/x(x²+1)^4d(x²+1)
=2∫1/(1+x²)^4dx
=2∫(π/4到π/2)1/(sec²v)^4dtanv
=1/4∫(cos2v+1)³dv
=1/8∫cos²2vdsin2v+3/4∫cos²2vdv+3/4∫cos2vdv+v/4
=(sin2v-sin³2v/3)/8+(3/8)(v+sin4v/4)+(3/8)sin2v+v/4
=5/8*π/4=5π/32
=∫(1到∞)1/x(x²+1)^4d(x²+1)
=2∫1/(1+x²)^4dx
=2∫(π/4到π/2)1/(sec²v)^4dtanv
=1/4∫(cos2v+1)³dv
=1/8∫cos²2vdsin2v+3/4∫cos²2vdv+3/4∫cos2vdv+v/4
=(sin2v-sin³2v/3)/8+(3/8)(v+sin4v/4)+(3/8)sin2v+v/4
=5/8*π/4=5π/32
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