Chapter 5, Systems of Equations and Inequalities - Section 5.3 - Partial Fractions - 5.3 Exercises - Page 462: 27

$\frac{x^2+1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{2}{x+1}$

$\frac{x^2+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$, $x^2+1=Ax(x+1)+B(x+1)+Cx^2$, $x^2+1=A(x^2+x)+B(x+1)+Cx^2$, $x^2+1=(A+C)x^2+(A+B)x+B=x^2+1$, $A+C=1$ $A+B=0$, $B=1$, $A+1=0, A=-1$ and $-1+C=1, C=2$. thus, $\frac{x^2+1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{2}{x+1}$