Question 1073127
{{{x^4+5x^2+6=(x^2+a)(x^2+b)}}}
Substitute,
{{{u=x^2}}}
So,
{{{1/(u^2+5u+6)=A/(u+2)+B/(u+3)}}}
{{{1=A(u+3)+B(u+2)}}}
{{{(A+B)u+(3A+2B)=1}}}
So,
{{{A+B=0}}}
{{{A=-B}}}
and
{{{3A+2B=1}}}
{{{-3B+2B=1}}}
{{{-B=1}}}
{{{B=-1}}}
and
{{{A=1}}}
So,
{{{1/(u^2+5u+6)=1/(u+2)-1/(u+3)}}}
{{{1/(x^4+5x^2+6)=1/(x^2+2)-1/(x^2+3)}}}