Answer
$$\ln \left| s \right| + \frac{9}{{{s^2} + 9}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{s^4} + 81}}{{s{{\left( {{s^2} + 9} \right)}^2}}}} ds \cr
& {\text{Decompose the integrand }}\frac{{{s^4} + 81}}{{s{{\left( {{s^2} + 9} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{{{s^4} + 81}}{{s{{\left( {{s^2} + 9} \right)}^2}}} = \frac{A}{s} + \frac{{Bs + C}}{{{s^2} + 9}} + \frac{{Ds + E}}{{{{\left( {{s^2} + 9} \right)}^2}}} \cr
& {\text{multiply by }}s{\left( {{s^2} + 9} \right)^2}{\text{ and simplify}} \cr
& {s^4} + 81 = A{\left( {{s^2} + 9} \right)^2} + \left( {Bs + C} \right)s\left( {{s^2} + 9} \right) + \left( {Ds + E} \right)s \cr
& {\text{Expanding the equation }} \cr
& {s^4} + 81 = A\left( {{s^4} + 18{s^2} + 81} \right) + \left( {Bs + C} \right)\left( {{s^3} + 9s} \right) + \left( {D{s^2} + Es} \right) \cr
& {s^4} + 81 = A{s^4} + 18A{s^2} + 81A + B{s^4} + 9B{s^2} + C{s^3} + 9Cs + D{s^2} + Es \cr
& \cr
& {\text{Group terms}} \cr
& {s^4} + 81 = \left( {A + B} \right){s^4} + C{s^3} + \left( {18A + 9B + D} \right){s^2} + 9Cs + Es + 81A \cr
& {\text{Equating coefficients}} \cr
& \,\,\,\,\,\,\,\,A + B = 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C = 0 \cr
& \,\,\,\,\,\,\,\,\,18A + 9B + D = 0 \cr
& \,\,\,\,\,\,\,\,9C + E = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,A = 1 \cr
& {\text{Solving the system of equations by using a calculator}}{\text{, we obtain}} \cr
& A = 1,\,\,\,B = 0,\,\,C = 0,\,\,\,D = - 18,\,\,\,E = 0 \cr
& \cr
& \frac{{{s^4} + 81}}{{s{{\left( {{s^2} + 9} \right)}^2}}} = \frac{A}{s} + \frac{{Bs + C}}{{{s^2} + 9}} + \frac{{Ds + E}}{{{{\left( {{s^2} + 9} \right)}^2}}} = \frac{1}{s} - \frac{{18s}}{{{{\left( {{s^2} + 9} \right)}^2}}} \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{{{s^4} + 81}}{{s{{\left( {{s^2} + 9} \right)}^2}}}} ds = \int {\left( {\frac{1}{s} - \frac{{18}}{{{{\left( {{s^2} + 9} \right)}^2}}}} \right)} ds \cr
& = \int {\frac{1}{s}} ds - 9\int {{{\left( {{s^2} + 9} \right)}^{ - 2}}\left( {2s} \right)ds} \cr
& {\text{integrating}} \cr
& = \ln \left| s \right| - 9\left( {\frac{{{{\left( {{s^2} + 9} \right)}^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = \ln \left| s \right| + \frac{9}{{{s^2} + 9}} + C \cr} $$
