∫ 1/x^2+4x＋5 dx=∫1/(x+2)²+1 dx=arctan(x+2)
所以，原式=∫<-∞,-2>1/x^2+4x＋5 dx+ ∫<-2,+∞>1/x^2+4x＋5 dx
=arctan(x+2)|<-∞,-2>+arctan(x+2)|<-2,+∞>
=0- (- π/2) + π/2- 0=π.
∫ 1/x^2+4x＋5 dx
=∫ 1/(x+2)^2＋1 dx
=arctan(x+2)（下限-∞，上限∞）=π/2-(-π/2)=π
原式=∫(-∞,+∞) d(x+2)/[(x+2)^2+1]
=arctan(x+2) |(-∞,+∞)
=π/2-(-π/2)
=π
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