∫(0,1)x²/(1+x²)³dx

设x=tant,则dx=sec²tdt
∵当x=0时,t=0
当x=1时,t=π/4
∴∫(0,1)x²/(1+x²)³dx=∫(0,π/4)tan²t*sec²tdt/(sec²t)³
=∫(0,π/4)sin²t*cos²tdt
=1/4∫(0,π/4)sin²(2t)dt
=1/8∫(0,π/4)[1-cos(4t)]dt
=1/8[t-sin(4t)/4]|(0,π/4)
=1/8[π/4-0-0+0]
=π/32.