Question

Express 1/(x^3-9x) as partial fraction

Answer 1

\[\frac{ 1 }{ x^3-9x }=\frac{ 1 }{x \left( x ^{2} -9\right) }=\frac{ 1 }{x \left( x+3 \right)\left( x-3 \right) }\]
\[\frac{ 1 }{ x(x+3)(x-3) }=\frac{ A }{x }+\frac{ B }{x+3 }+\frac{ C }{x-3 }\]
1=A(x+3)(x-3)+Bx(x-3)+Cx(x+3)
plug x=0,3,-3
and get the values of A,B,C