Answer
$$y - \ln \left| y \right| = {\left( {x + 1} \right)^3} + C$$
Work Step by Step
$$\eqalign{
& y' = \frac{{3y{{\left( {x + 1} \right)}^2}}}{{y - 1}} \cr
& \frac{{dy}}{{dx}} = \frac{{3y{{\left( {x + 1} \right)}^2}}}{{y - 1}} \cr
& {\text{Separating variables gives}} \cr
& \frac{{y - 1}}{y}dy = 3{\left( {x + 1} \right)^2}dx \cr
& \left( {1 - \frac{1}{y}} \right)dy = 3{\left( {x + 1} \right)^2}dx \cr
& {\text{integrating}} \cr
& \int {\left( {1 - \frac{1}{y}} \right)} dy = \int {3{{\left( {x + 1} \right)}^2}} dx \cr
& y - \ln \left| y \right| = 3\left( {\frac{{{{\left( {x + 1} \right)}^3}}}{3}} \right) + C \cr
& y - \ln \left| y \right| = {\left( {x + 1} \right)^3} + C \cr} $$
