Answer
$$y = \left( {t + 1} \right) + \frac{{\ln \left( {t + 1} \right)}}{{{{\left( {t + 1} \right)}^2}}} + \frac{C}{{{{\left( {t + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \left( {t + 1} \right)\frac{{ds}}{{dt}} + 2s = 3\left( {t + 1} \right) + \frac{1}{{{{\left( {t + 1} \right)}^2}}},\,\,\,\,\,\,\,\,\,\,\,\,\,t > - 1 \cr
& {\text{Divide each term of the differential equation by }}t + 1 \cr
& \frac{{\left( {t + 1} \right)}}{{t + 1}}\frac{{ds}}{{dt}} + \frac{2}{{t + 1}}s = \frac{{3\left( {t + 1} \right)}}{{t + 1}} + \frac{1}{{{{\left( {t + 1} \right)}^3}}} \cr
& \frac{{ds}}{{dt}} + \frac{2}{{t + 1}}s = 3 + \frac{1}{{{{\left( {t + 1} \right)}^3}}}\,\,\,\,\,\,\,\left( 1 \right) \cr
& {\text{The equation is written in the form }}\frac{{ds}}{{dt}} + P\left( t \right)s = Q\left( t \right) \cr
& P\left( t \right) = \frac{2}{{t + 1}},\,\,\,Q\left( t \right) = 3 + \frac{1}{{{{\left( {t + 1} \right)}^3}}} \cr
& {\text{The integrating factor is }}\cr
& v\left( t \right) = {e^{\int {\frac{2}{{t + 1}}} dt}} = {e^{2\ln \left( {t + 1} \right)}} = {\left( {t + 1} \right)^2} \cr
& {\text{Multiply equation 1 by the integrating factor}} \cr
& {\left( {t + 1} \right)^2}\frac{{ds}}{{dt}} + {\left( {t + 1} \right)^2}\frac{2}{{t + 1}}s = 3{\left( {t + 1} \right)^2} + {\left( {t + 1} \right)^2}\frac{1}{{{{\left( {t + 1} \right)}^3}}} \cr
& {\left( {t + 1} \right)^2}\frac{{ds}}{{dt}} + 2\left( {t + 1} \right)s = 3{\left( {t + 1} \right)^2} + \frac{1}{{t + 1}} \cr
& \cr
& {\text{Write the left - hand in the form }}\frac{d}{{ds}}\left[ {y{{\left( {t + 1} \right)}^2}} \right] \cr
& \frac{d}{{ds}}\left[ {y{{\left( {t + 1} \right)}^2}} \right] = 3{\left( {t + 1} \right)^2} + \frac{1}{{t + 1}} \cr
& \cr
& {\text{Integrate both sides of the differential equation}} \cr
& y{\left( {t + 1} \right)^2} = \int {3{{\left( {t + 1} \right)}^2}} ds + \int {\frac{1}{{t + 1}}} ds \cr
& y{\left( {t + 1} \right)^2} = 3\left( {\frac{{{{\left( {t + 1} \right)}^3}}}{3}} \right) + \ln \left| {t + 1} \right| + C \cr
& y{\left( {t + 1} \right)^2} = {\left( {t + 1} \right)^3} + \ln \left( {t + 1} \right) + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{{{\left( {t + 1} \right)}^3}}}{{{{\left( {t + 1} \right)}^2}}} + \frac{{\ln \left( {t + 1} \right)}}{{{{\left( {t + 1} \right)}^2}}} + \frac{C}{{{{\left( {t + 1} \right)}^2}}} \cr
& y = \left( {t + 1} \right) + \frac{{\ln \left( {t + 1} \right)}}{{{{\left( {t + 1} \right)}^2}}} + \frac{C}{{{{\left( {t + 1} \right)}^2}}} \cr} $$
