用换元法求定积分∫dx/(x^2+2x+2) ∫上面为0,下面为-2

问题描述:

用换元法求定积分
∫dx/(x^2+2x+2) ∫上面为0,下面为-2

令a=x+1
则x=0,a=1
x=-2,a=-1
dx=da
x²+2x+2=x²+2x+1+1=a²+1
原式=∫(1,-1)da/(a²+1)
=arctana(1,-1)
=π/4-(-π/4)
=π/2