A calculus problem by Hamza A

Use substitution $u=x-1$ :

$\int_{1}^{2}x \ln{(x-1)} dx=\int_{0}^{1}(u+1)\ln{(u)}du=\left [ \frac{1}{2}u^2 \ln{(u)}-\frac{1}{4} u^2+u \ln{(u)}-u \right ]_{0}^{1}=(-\frac{5}{4}-0)=-\frac{5}{4}$

$\begin{aligned} I & = \int_1^2 x \ln(x-1) \space dx \quad \quad \small \color{#3D99F6}{\text{Let }u = \ln (x-1) \space \Rightarrow e^u = x - 1 \space \Rightarrow x = 1+e^u \space \Rightarrow dx = e^u \space du} \\ & = \int_{-\infty}^0 (1+e^u) u e^u \space du \quad \quad \small \color{#3D99F6}{\text{Let }u = -t \space \Rightarrow du = - \space dt} \\ & = \int_{-\infty}^0 t e^{-t} (1+e^{-t}) \space dt \\ & = - \int^{\infty}_0 (t e^{-t} + te^{-2t}) \space dt \\ & = - \int^{\infty}_0 t^{2-1} e^{-t} \space dt - \frac{1}{4} \int^{\infty}_0 (2t)^{2-1} e^{-2t} \space d(2t) \\ & = - \Gamma (2) - \frac{1}{4} \Gamma (2) = - \frac{5}{4} \Gamma (2) = - \frac{5}{4} \end{aligned}$

$\Rightarrow a + b = 5 + 4 = \boxed{9}$