## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{1}{t^2+9}$
$\frac{t-3}{t^2+9}\times\frac{t+3}{t^2-9}$ Factorise the fraction $\frac{t-3}{t^2-9}$: $=\frac{t-3}{t^2+9}\times\frac{t+3}{(t-3)(t+3)}$ Multiply the fractions: $=\frac{(t-3)(t+3)}{(t^2+9)(t-3)(t+3)}$ Simplify the fraction: $=\frac{1}{t^2+9}$