Answer
$$w = {\left( {\frac{3}{2}\ln \left| x \right| - \frac{1}{{2x}} + C} \right)^2}$$
Work Step by Step
$$\eqalign{
& {x^2}\frac{{dw}}{{dx}} = \sqrt w \left( {3x + 1} \right) \cr
& {\text{separate}} \cr
& x\frac{{dw}}{{\sqrt w }} = \frac{{3x + 1}}{{{x^2}}}dx \cr
& {\text{integrate both sides}} \cr
& \int {{w^{ - 1/2}}dw} = \int {\left( {\frac{{3x}}{{{x^2}}} + \frac{1}{{{x^2}}}} \right)} dx \cr
& \frac{{{w^{1/2}}}}{{1/2}} = 3\ln \left| x \right| - \frac{1}{x} + C \cr
& 2{w^{1/2}} = 3\ln \left| x \right| - \frac{1}{x} + C \cr
& {\text{solve for }}w \cr
& {w^{1/2}} = \frac{3}{2}\ln \left| x \right| - \frac{1}{{2x}} + C \cr
& w = {\left( {\frac{3}{2}\ln \left| x \right| - \frac{1}{{2x}} + C} \right)^2} \cr} $$
