Question 1085784
{{{2/(x(x+1)(x+2))=A/x+B/(x+1)+C/(x+2)}}}
{{{2=A(x+1)(x+2)+Bx(x+2)+Cx(x+1)}}}
{{{2=A(x^2+3x+2)+Bx^2+2Bx+Cx^2+Cx}}}
{{{2=(A+B+C)x^2+(3A+2B+C)x+2A}}}
So then,
{{{2A=2}}}
{{{A=1}}}
which leaves
{{{1+B+C=0}}}
{{{3+2B+C=0}}}
Subtract the first equation from the second,
{{{3+2B+C-1-B-C=0}}}
{{{2+B=0}}}
{{{B=-2}}}
and finally,
{{{1-2+C=0}}}
{{{-1+C=0}}}
{{{C=1}}}
So,
{{{2/(x(x+1)(x+2))=1/x-2/(x+1)+1/(x+2)}}}