解
∫x/(x²+2x+2)dx
=1/2∫(2x+2-2)/(x²+2x+2)dx
=1/2∫(2x+2)/(x²+2x+2)dx-∫1/(x²+2x+2)dx
=1/2∫1/(x²+2x+2)d(x²+2x+2)-∫1/[(x+1)²+1]dx
=1/2∫1/udu-∫1/[(x+1)²+1]d(x+1)
=1/2ln|u|-∫1/(u²+1)du
=1/2ln(x²+2x+2)-acrtanu+C
=1/2ln(x²+2x+2)-arctan(x+1)+C