Precalculus, Chapter 7, 7.4, Section 7.4, Problem 42

You need to decompose the fraction in simple irreducible fractions, such that:
(x+1)/(x^3(x^2+1)^2) = A/x + B/(x^2) + C/(x^3) + (Dx+E)/(x^2+1) + (Fx+G)/((x^2+1)^2)
You need to bring to the same denominator all fractions, such that:
x+1= A(x^2(x^2+1)^2) + Bx(x^2+1)^2 + C(x^2+1)^2 + (Dx+E)x^3*(x^2+1) + (Fx+G)*x^3
x+1= Ax^2(x^4 + 2x^2 + 1) + Bx(x^4 + 2x^2 + 1) + C(x^4 + 2x^2 + 1) + (Dx^4+Ex^3)*(x^2+1) + Fx^4 + Gx^3
x+1= Ax^6+ + 2Ax^4 + Ax^2 +Bx^5 + 2Bx^3 + Bx+ Cx^4 + 2Cx^2 +C + Dx^6 + Dx^4 + Ex^5 + Ex^3 + Fx^4 + Gx^3
You need to group the terms having the same power of x:
x+1=x^6(A+D) + x^5(B+E) + x^4(2A+C+D+F) + x^3(2B+E+G) + x^2(A+2C) + x(B) + C
Comparing the expressions both sides yields:
A+D = 0
B+E = 0
2A+C+D+F = 0 => F = 1
2B+E+G = 0 => G = -1
A+2C = 0 => A = -2 => D = 2
B= 1 => E = -1
C = 1
Hence, the partial fraction decomposition of the improper rational expression is (x+1)/(x^3(x^2+1)^2) = -2/x + 1/(x^2) + 1/(x^3) + (2x-1)/(x^2+1) + (x-1)/((x^2+1)^2)