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

$$\frac{t^4+9}{t^4+9t^2}=1+\frac{1}{t^2}-\frac{10}{t^2+9}$$
$$\frac{t^4+9}{t^4+9t^2}$$ Since the degree of the numerator is not lower than that of the denominator, we need to divide $t^4+9$ by $t^4+9t^2$ to get the remainder first, which gives us $$\frac{t^4+9}{t^4+9t^2}=1+\frac{(-9t^2)+9}{t^4+9t^2}=1+\frac{9-9t^2}{t^2(t^2+9)}$$ Now we use partial fraction method for the remainder. $$\frac{9-9t^2}{t^2(t^2+9)}=\frac{A}{t}+\frac{B}{t^2}+\frac{Ct+D}{t^2+9}$$ Here, notice that $t^2+9$ is an irreducible quadratic factor, as it has no real roots, so its faction should be in the form $Ct+D$. To find $A$, $B$, $C$ and $D$, we clear fractions and get $$9-9t^2=At(t^2+9)+B(t^2+9)+(Ct+D)t^2$$ $$9-9t^2=At^3+9At+Bt^2+9B+Ct^3+Dt^2$$ $$9-9t^2=t^3(A+C)+t^2(B+D)+9At+9B$$ Equating coefficients of corresponding powers of $t$, we get $9A=0$, so $A=0$ $9B=9$, so $B=1$ $A+C=0$. As $A=0$, we have $C=0$ $B+D=-9$. As $B=1$, we have $D=-10$ Therefore, $$\frac{t^4+9}{t^4+9t^2}=1+\frac{9-9t^2}{t^4+9t^2}=1+\frac{1}{t^2}-\frac{10}{t^2+9}$$