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∫(1/x^2+4x+5)dx定积分范围是上面-3下面-1

题目详情
∫(1/x^2+4x+5)dx定积分范围是上面-3下面-1
▼优质解答
答案和解析
∫dx/(x^2+4x+5)
=∫d(x+2)/[(x+2)^2+1]
=arctan(x+2)+C
∫[-1,-3]dx/(x^2+4x+5)
=arctan(x+2)|[-1,-3]
=arctan(-1)-arctan1
=-π/2