令x=tanu，则u=arctanx∫[1/(1+x²)²]dx=∫[1/(1+tan²u)²]d(tanu)=∫(sec²u/sec⁴u)du=∫cos²udu=¼∫(1+cos2u)d(2u)=¼(2u+sin2u) +C=½(u+sinucosu) +C=½(arctanx +x/(x²+1)] +C
