∫<-∞,+∞>dx/[(2x+1)^2+2^2]=(1/4)∫<-∞,+∞>dx/[(x+1/2)^2+1](恒等变形)=(1/4)arctan(x+1/2)|<-∞,+∞>(用公式)=π/4.(反正切)
