[tex]\begin{gathered} f^{\prime\prime}(x)=-\frac{4}{(x-1)^2}-2 \\ u=x-1 \\ f^{\prime}(x)=\int_{x>1}(-\frac{4}{(x-1)^2}-2)\cdot dx=-4\int_{u>0}u^{-2}\cdot du-2\int_{x>1}dx+c \\ f^{\prime}(x)=\frac{4}{u}-2x \\ f^{\prime}(x)=\frac{4}{x-1}-2x+c \\ f^{\prime}(x)=\frac{-2x^2+2x+4}{x-1}+c \\ f^{\prime}(2)=0 \\ c=-\frac{(-2\cdot2^2+2\cdot2+4)}{2-1} \\ c=0 \\ \begin{equation*} f^{\prime}(x)=\frac{-2x^2+2x+4}{x-1} \end{equation*} \\ f(x)=\int_{x>1}\frac{-2x^{2}+2x+4}{x-1}\cdot dx \\ f(x)=-2\int_{x>1}\frac{x^2}{x-1}dx+2\int_{x>1}\frac{x}{x-1}dx+4\int\frac{1}{x-1}dx \\ f(x)=-(x^2+2x+2\ln|x-1|-3)+2(x-1+\ln|x-1|)+2\ln|x-1|+c \\ f(x)=-x^2+2\ln|x-1|+1+c \\ f(2)=3 \\ -2^2+2\ln|2-1|+1+c=3 \\ c=6 \\ \therefore f(x)=-x^2+2\ln|x-1|+7 \end{gathered}[/tex]
