## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{t^2+8}{t^2-5t+6}=1-\frac{12}{t-2}+\frac{17}{t-3}$$
$$\frac{t^2+8}{t^2-5t+6}$$ Since the degree of the numerator is not lower than that of the denominator, we need to divide $t^2+8$ by $t^2-5t+6$ to get the remainder first. Dividing them gives us $$\frac{t^2+8}{t^2-5t+6}=1+\frac{5t+2}{t^2-5t+6}=1+\frac{5t+2}{(t-2)(t-3)}$$ Now we use partial fraction method for the remainder: $$\frac{5t+2}{(t-2)(t-3)}=\frac{A}{t-2}+\frac{B}{t-3}$$ To find $A$ and $B$, we clear fractions and get $$5t+2=A(t-3)+B(t-2)$$ $$5t+2=At-3A+Bt-2B$$ $$5t+2=t(A+B)+(-3A-2B)$$ Equating coefficients of corresponding powers of $t$, we get - $A+B=5$ - $-3A-2B=2$ So $A=-12$ and $B=17$ Therefore, $$\frac{t^2+8}{t^2-5t+6}=1+\frac{5t+2}{t^2-5t+6}=1-\frac{12}{t-2}+\frac{17}{t-3}$$