广义积分（下限-∞,上限∞）∫ 1/x^2+4x＋5 dx怎么算呢?
网友回答
∫ 1/x^2+4x＋5 dx
=∫ 1/(x+2)^2＋1 dx
=arctan(x+2)（下限-∞,上限∞）=π/2-(-π/2)=π
======以下答案可供参考======
供参考答案1:
原式=∫(-∞,+∞) d(x+2)/[(x+2)^2+1]
=arctan(x+2) |(-∞,+∞)
=π/2-(-π/2)
=π供参考答案2:
∫ 1/x^2+4x＋5 dx=∫1/(x+2)²+1 dx=arctan(x+2)
所以，原式=∫1/x^2+4x＋5 dx+ ∫1/x^2+4x＋5 dx
=arctan(x+2)|+arctan(x+2)|
=0- (- π/2) + π/2- 0=π.