Answer
$\displaystyle \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
Work Step by Step
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,
$B=-1$
$\displaystyle \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{(-1)x+0}{(x^{2}+1)}$
$\displaystyle \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$
