广义积分(下限-∞,上限∞)∫ 1/x^2+4x+5 dx怎么算呢?
问题描述:
广义积分(下限-∞,上限∞)∫ 1/x^2+4x+5 dx怎么算呢?
答
∫ 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.
答
原式=∫(-∞,+∞) d(x+2)/[(x+2)^2+1]
=arctan(x+2) |(-∞,+∞)
=π/2-(-π/2)
=π
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