9.(2023春•朝阳期末)已知函数f(x)=ln(x−1)−a2x(a∈R)f(x)=ln(x-1)-a^{2}x(a\in R)f(x)=ln(x−1)−a2x(a∈R).(1)讨论函数f(x)f(x)f(x)的单调区间;(2)若函数f(x)f(x)f(x)在x=2x=2x=2处取得极值,对∀x∈(1,+∞)\forall x\in (1,+\infty )∀x∈(1,+∞),f(x)⩽bx+lnx−1x+1f(x)\leqslant bx+ln\frac{x-1}{x}+1f(x)⩽bx+lnxx−1+1恒成立,求实数bbb的取值范围.