Answer
$$1 + \frac{1}{{{t^2}}} - \frac{{10}}{{{t^2} + 9}}$$
Work Step by Step
$$\eqalign{
& \frac{{{t^4} + 9}}{{{t^4} + 9{t^2}}} \cr
& {\text{Divide the denominator into the numerator to get a poynomial plus a }} \cr
& {\text{proper fraction}} \cr
& \frac{{{t^4} + 9}}{{{t^4} + 9{t^2}}} = 1 + \frac{{9 - 9{t^2}}}{{{t^4} + 9{t^2}}} \cr
& {\text{decomposing }}\frac{{9 - 9{t^2}}}{{{t^4} + 9{t^2}}}{\text{ into partial fractions}} \cr
& \frac{{9 - 9{t^2}}}{{{t^4} + 9{t^2}}} = \frac{{9 - 9{t^2}}}{{{t^2}\left( {{t^2} + 9} \right)}} \cr
& {\text{The partial fraction decomposition has the form}} \cr
& \frac{{9 - 9{t^2}}}{{{t^2}\left( {{t^2} + 9} \right)}} = \frac{A}{t} + \frac{B}{{{t^2}}} + \frac{{Ct + D}}{{{t^2} + 9}}\,\,\,\left( {\bf{1}} \right) \cr
& {\text{Multiplying by }}{t^2}\left( {{t^2} + 9} \right){\text{, we have}} \cr
& 9 - 9{t^2} = At\left( {{t^2} + 9} \right) + B\left( {{t^2} + 9} \right) + \left( {Ct + D} \right){t^2} \cr
& 9 - 9{t^2} = A{t^3} + 9At + B{t^2} + 9B + C{t^3} + D{t^2} \cr
& 9 - 9{t^2} = \left( {A{t^3} + C{t^3}} \right) + \left( {B{t^2} + D{t^2}} \right) + 9At + 9B \cr
& {\text{If we equate coefficients}}{\text{, we get the system}} \cr
& A + C = 0,\,\,\,\,\,\,\,\,\,\,\,\,B + D = - 9,\,\,\,\,\,\,\,\,\,\,\,\,\,9A = 0,\,\,\,\,\,\,9B = 9 \cr
& {\text{Solving, we get}} \cr
& A = 0,\,\,C = 0,\,\,\,B = 1,\,\,\,\,D = - 10 \cr
& {\text{Then substituting }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \frac{{9 - 9{t^2}}}{{{t^2}\left( {{t^2} + 9} \right)}} = \frac{1}{{{t^2}}} - \frac{{10}}{{{t^2} + 9}} \cr
& \cr
& \frac{{{t^4} + 9}}{{{t^4} + 9{t^2}}} = 1 + \frac{1}{{{t^2}}} - \frac{{10}}{{{t^2} + 9}} \cr} $$
