为自己加油

计算∫_-∞^+∞[1/(x²+2x+2)]dx．

∫_+∞^-∞[1/(x²+2x+2)]dx =lim_α→-∞∫_α⁰[d(x+1)]/[1+(x+1)²]+lim_b→+∞∫₀^b[d(x+1)]/[1+(x+1)²] =lim_α→-∞arctan(x+1)|_α⁰+lim_b→+∞arctan(x+1)|₀^b =lim_α→-∞(arctanl-arctanα)+lim_b→+∞(arctanb-arctanl) =π/4-(-π/2)+π/2-π/4=π