显然分母=(x+1)(x^2+1),设原分式=A/(x+1)+(Bx+C)/(x^2+1)
于是通分得[(A+B)x^2+(B+C)x+A+C]/(x^3+x^2+x+1),于是系数相等得
A+B=0,B+C=1,A+C=7,三式相加/2得A+B+C=4,所以A=3,B=-3,C=4
所以原式可表示为3/(x+1)+(-3x+4)/(x^2+1)
(x+7)/(x^3+x^2+x+1)
=(x+7)/[x^2(x+1)+(x+1)]
=(x+7)/[(x+1)(x^2+1)]
=(x+7)/(x^2+1)+(x+7)/(x+1)
所求的两个分式分别为(x+7)/(x^2+1)和(x+7)/(x+1)