\(((x+2)y'+4y)/((x+2))=((3x+6)^{-2} \ln x)/((x+2))\)

\(y'+(4/(x+2))y= \ln x/(9(x+2)^3)\) (1)

Now, we know that solution of differential equation\((y'+p(x)y= q(x))\) is given by

\(y(I.F.)= \int q(x)(I.F.)dx\) (2)

where \(I.F.= e^{\int p(x)dx}\)

So, finding integrating factor of equation (1)

\(I.F.= e^{\int 4/(x+2)dx}\)

\(= e^{4 \int dx/(x+2)}\) \(( :' \int dx/(x+a)= \ln(x+a)+c)\)

\(= e^{4(\ln(x+2))}\)

\(= e^{\ln (x+2)^4}\)

\(= (x+2)^4\)

So, solution of given differential equation will be

\(y(x+2)^4= \int \ln x/(9(x+2)^3) (x+2)^4 dx\)

\(y(x+2)^4= 1/9 \int (x+2) \ln xdx\)

\(y(x+2)^4= 1/9 [\ln x \int(x+2)dx- \int 1/x [\int(x+2)dx]dx]\)

\(= 1/9 [((x+2)^2 \ln x)/2 - \int 1/x ((x+2)^2)/2 dx]\)

\(= ((x+2)^2 \ln x)/18-1/2 \int (x^2+4x+4)/x dx\)

\(= ((x+2)^2 \ln x)/18-1/2(x^2/2+4x+4\ln x)+c\)