## Thomas' Calculus 13th Edition

$$1 + \frac{{17}}{{t - 3}} - \frac{{12}}{{t - 2}}$$
\eqalign{ & \frac{{{t^2} + 8}}{{{t^2} - 5t + 6}} \cr & {\text{Divide the denominator into the numerator to get a}} \cr & {\text{polynomial with a proper fraction}} \cr & \frac{{{t^2} + 8}}{{{t^2} - 5t + 6}} = 1 + \frac{{5t + 2}}{{{t^2} - 5t + 6}} \cr & {\text{decomposing }}\frac{{5t + 2}}{{{t^2} - 5t + 6}}{\text{ into partial fractions}} \cr & \frac{{5t + 2}}{{{t^2} - 5t + 6}} = \frac{{5t + 2}}{{\left( {t - 3} \right)\left( {t - 2} \right)}} \cr & {\text{The partial fraction decomposition has the form}} \cr & \frac{{5t + 2}}{{\left( {t - 3} \right)\left( {t - 2} \right)}} = \frac{A}{{t - 3}} + \frac{B}{{t - 2}}\,\,\,\left( {\bf{1}} \right) \cr & {\text{Multiplying by }}\left( {t - 3} \right)\left( {t - 2} \right){\text{, we have}} \cr & 5t + 2 = A\left( {t - 2} \right) + B\left( {t - 3} \right) \cr & \cr & \,\,\,\,\,{\text{set }}t = 3 \cr & 5\left( 3 \right) + 2 = A\left( {3 - 2} \right) + B\left( 0 \right) \cr & A = 17 \cr & \cr & \,\,\,\,\,{\text{set }}t = 2 \cr & 5\left( 2 \right) + 2 = A\left( 0 \right) + B\left( {2 - 3} \right) \cr & B = - 12 \cr & \cr & {\text{Then substituting }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr & \frac{{5t + 2}}{{\left( {t - 3} \right)\left( {t - 2} \right)}} = \frac{{17}}{{t - 3}} + \frac{{ - 12}}{{t - 2}} \cr & {\text{Then}}{\text{,}} \cr & \frac{{{t^2} + 8}}{{{t^2} - 5t + 6}} = 1 + \frac{{5t + 2}}{{{t^2} - 5t + 6}} \cr & \frac{{{t^2} + 8}}{{{t^2} - 5t + 6}} = 1 + \frac{{17}}{{t - 3}} - \frac{{12}}{{t - 2}} \cr}