广义积分∫ [1/(x^2+4x+5)]dx = .
问题描述:
广义积分∫ [1/(x^2+4x+5)]dx = .
答
∫ [1/(x²+4x+5)]dx
= ∫ 1/[(x+2)²+1]d(x+2) + ∫ 1/[(x+2)²+1]d(x+2)
=arctan(x+2)| +arctan(x+2)|
=π/2-0+0-(-π/2)
= π