## College Algebra (10th Edition)

$\displaystyle \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
case 3: Q contains a nonrepeated irreducible quadratic factor. $\displaystyle \frac{1}{x(x^{2}+1)}=\frac{A}{x}+\frac{Bx+C}{(x^{2}+1)}$ ... write the RHs with a common denominator $\displaystyle \frac{1}{x(x^{2}+1)}=\frac{A(x^{2}+1)+x(Bx+C)}{x(x^{2}+1)}$ ... equate the numerators $1=Ax^{2}+A+Bx^{2}+Cx$ $1=(A+B)x^{2}+ Cx+A$ ... equate the coefficients of the polynomials on the LHS and RHS: $\left\{\begin{array}{llll} A & +B & & =0\\ & & C & =0\\ A & & & =1 \end{array}\right.$ So, $A=1,\ C=0$, and by substituting into the first equation, $B=-1$ $\displaystyle \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{(-1)x+0}{(x^{2}+1)}$ $\displaystyle \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$