∫[-1,-3][1/(x^2+4x+5)]dx定积分
问题描述：
∫[-1,-3][1/(x^2+4x+5)]dx定积分
答
∫[-1,-3][1/(x^2+4x+5)]dx
=∫[-1,-3][1/[(x+2)²+1]d(x+2)
=arctan(x+2)[-1,-3]
是不是-1是上限?
=arctan1-arctan(-1)
=2arctan1
=π/2是下限哦，那就是相反数