Question

求1/x^2+4x+8上限是正无群下限是0的定积分

Answer 1

F(x)=∫dx/(x^2+4x+8)

=∫dx/[(x+2)^2+4]

=∫dx/4[(x/2+1)^2+1]

=(1/2)∫d(x/2+1)/[(x/2+1)^2+1]

=(1/2)arctan(x/2+1)+C

代入上下限,可知

F(+∞)=π/4+C

F(0)=π/8+C

故定积分=π/8