## Thomas' Calculus 13th Edition

$$\frac{2}{{x - 3}} + \frac{3}{{x - 2}}$$
\eqalign{ & \frac{{5x - 13}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} \cr & {\text{The partial fraction decomposition has the form}} \cr & \frac{{5x - 13}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \frac{A}{{x - 3}} + \frac{B}{{x - 2}}\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{Find the values of the coefficients }}A,B{\text{; we clear fractions to obtain}} \cr & 5x - 13 = A\left( {x - 2} \right) + B\left( {x - 3} \right) \cr & \,\,\,\,\,{\text{set }}x = 3 \cr & 5\left( 3 \right) - 13 = A\left( {3 - 2} \right) + B\left( {3 - 3} \right) \cr & 2 = A \cr & \,\,\,\,\,{\text{set }}x = 2 \cr & 5\left( 2 \right) - 13 = A\left( {2 - 2} \right) + B\left( {2 - 3} \right) \cr & - 3 = B\left( { - 1} \right) \cr & B = 3 \cr & {\text{Then substitute }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr & \frac{{5x - 13}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \frac{2}{{x - 3}} + \frac{3}{{x - 2}} \cr}