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积分:(x^2+2x+2)^(-2)dx
=积分:dx/[(x+1)^2+1]^2
=积分;d(x+1)/[(x+1)^2+1]^2
令x+1=t
=积分:dt/[t^2+1]^2
=积分;(t^2+1-t^2)dt/[(t^2+1)]^2
=积分:dt/(t^2+1)-积分:t^2dt/(t^2+1)
=积分:dt/(t^2+1)+1/2积分:td(1/(t^2+1))
=积分:dt/(t^2+1)+1/2[t/(t^2+1)-积分:dt/(t^2+1)]
=1/2积分;dt/(t^2+1)+t/2(t^2+1)
=1/2*arctant+t/2(t^2+1)+C
将x+1=t代入有:
=1/2arctan(x+1)+(x+1)/2(x+1)^2+1)+C

S是积分号
S(x^2+2x+2)^(-2)dx
=S{(x+1)^2+1}^(-2)d(x+1)
令t=x+1
=S(t^2+1)^(-2)dt

设Y(1)=S(t^2+1)^(-1)dt=arctant+c
对于n>=2:
Y(n)=S(t^2+1)^(-n)dt
=S(t^2+1-t^2)/(t^2+1)dt
=S(t^2+1)^[-(n-1)]dt-St^2/(t^2+1)^(-n)dt
=S(t^2+1)^[-(n-1)]dt+[2(n-1)]^(-1)Std[(x^2+1)^(-1)]
=Y(n-1)+t/[2(n-1)(x^2+1)^(n-1)]-Y(n-1)/2(n-1)
于是得到递推关系：
Y(n)=(2n-3)Y(n-1)/2(n-1)+t/[2(n-1)(x^2+1)^(n-1)]
Y(1)=arctant+c
将n=2代入上递推关系得到：
1/2*{arctan(x+1)+(x+1)/[(x+1)^2+1]}

∫(x^2+2x+2)^(-2)dx
=∫[(x+1)^2+1]^(-2)dx
=∫[(x+1)^2+1]^(-2)d(x+1)
=∫d(x+1)/[(1+x)^2+1]
用arctanx的公式
得原式=arctan(x+1)+C

1/2 ((1 + x)/(2 + 2 x + x^2) + ArcTan[1 + x])

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