Question

求定积分!如图

如图

Answer 1

原式=∫(0,-2)(x+1)/[(x+1)^2+1]+1/[(x+1)^2+1]dx
=∫(0,-2)(x+1)/[(x+1)^2+1]dx+∫(0,-2)1/[(x+1)^2+1]dx
=1/2∫(0,-2)1/[(x+1)^2+1]d(x+1)^2+∫(0,-2)1/[(x+1)^2+1]d(x+1)
=1/2(0,-2)ln[x+1)^2+1]+(0,-2)arctan(x+1)
=π/4-(-π/4)
=π/2

Answer 2

原式=1/2∫(-2,0)d(x+1)^2/[(x+1)^2+1]+∫(-2,0)d(x+1)/[(x+1)^2+1]
=1/2*ln((x+1)^2+1)|(-2,0)+arctan(x+1)|(-2,0)
=1/2ln2-1/2ln2+arctan1-arctan(-1)
=π/4+π/4
=π/2

追问

加好友，我想问详细点，会加分的

追答

已加,其实做完后我发现还有更简单的方法,令
x+1=t
则
原式=∫(-1,1)(t+1)/(t^2+1)dt
=∫(-1,1)t/(t^2+1)dt+∫(-1,1)1/(t^2+1)dt
利用偶倍奇零,得
原式=2∫(0,1)1/(t^2+1)dt
=2arctant|(0,1)
=2*π/4
=π/2