Alg5.1.1

Find the partial fraction decomposition of $\frac{3}{x^2-9}\,$ .

Factor $x^2-9\,$ as $(x+3)(x-3)\,$ .

Then

$\frac{3}{(x+3)(x-3)} = \frac{A}{x+3}+\frac{B}{x-3}\,$

Multiply all terms on both sides by the common denominator $(x+3)(x-3)\,$ .

$3 = A(x-3) + B(x+3)\,$

So if $x=3\,$

then $3=6B\,$ .

Therefore $B=1/2\,$ .

Similarly if $x=-3\,$

then $3=-6A\,$ .

Therefore $A=-1/2\,$ .

So $\frac{3}{x^2-9} = \frac{-1/2}{x+3} + \frac{1/2}{x-3}\,$