Answer
$$y = \frac{{{t^3}}}{{3{{\left( {t - 1} \right)}^4}}} - \frac{t}{{{{\left( {t - 1} \right)}^4}}} + \frac{C}{{{{\left( {t - 1} \right)}^4}}}$$
Work Step by Step
$$\eqalign{
& {\left( {t - 1} \right)^3}\frac{{ds}}{{dt}} + 4{\left( {t - 1} \right)^2}s = t + 1,\,\,\,\,\,\,\,\,\,\,\,\,\,t > 1 \cr
& {\text{Divide each term of the differential equation by }}{\left( {t - 1} \right)^3} \cr
& \frac{{{{\left( {t - 1} \right)}^3}}}{{{{\left( {t - 1} \right)}^3}}}\frac{{ds}}{{dt}} + 4\frac{{{{\left( {t - 1} \right)}^2}s}}{{{{\left( {t - 1} \right)}^3}}} = \frac{{t + 1}}{{{{\left( {t - 1} \right)}^3}}} \cr
& \frac{{ds}}{{dt}} + \frac{4}{{t - 1}}s = \frac{{t + 1}}{{{{\left( {t - 1} \right)}^3}}}\,\,\,\,\,\left( 1 \right) \cr
& {\text{The equation is written in the form }}\frac{{ds}}{{dt}} + P\left( t \right)s = Q\left( t \right) \cr
& P\left( t \right) = \frac{4}{{t - 1}},\,\,\,Q\left( t \right) = \frac{{t + 1}}{{{{\left( {t - 1} \right)}^3}}} \cr
& {\text{The integrating factor is }}v\left( t \right) = {e^{\int {\frac{4}{{t - 1}}} dt}} = {e^{4\ln \left( {t - 1} \right)}} = {\left( {t - 1} \right)^4} \cr
& {\text{Multiply the differential }}\left( 1 \right){\text{ equation by the integrating factor}} \cr
& {\left( {t - 1} \right)^4}\frac{{ds}}{{dt}} + \frac{4}{{t - 1}}{\left( {t - 1} \right)^4}s = \frac{{t + 1}}{{{{\left( {t - 1} \right)}^3}}}{\left( {t - 1} \right)^4} \cr
& {\left( {t - 1} \right)^4}\frac{{ds}}{{dt}} + 4{\left( {t - 1} \right)^3}s = \left( {t + 1} \right)\left( {t - 1} \right) \cr
& \cr
& {\text{Write the left - hand in the form }}\frac{d}{{ds}}\left[ {y{{\left( {t - 1} \right)}^4}} \right] \cr
& \frac{d}{{ds}}\left[ {y{{\left( {t - 1} \right)}^4}} \right] = {t^2} - 1 \cr
& \cr
& {\text{Integrate both sides of the differential equation}} \cr
& y{\left( {t - 1} \right)^4} = \int {\left( {{t^2} - 1} \right)} dt \cr
& y{\left( {t - 1} \right)^4} = \frac{{{t^3}}}{3} - t + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{{t^3}}}{{3{{\left( {t - 1} \right)}^4}}} - \frac{t}{{{{\left( {t - 1} \right)}^4}}} + \frac{C}{{{{\left( {t - 1} \right)}^4}}} \cr} $$
