Limit of $\lim_{t\to 1^+} \int_t^{t^2} \frac{\arctan(x)}{x-1} dx$

Write $t = 1+\epsilon$ and notice that

\begin{align*} \int_{t}^{t^2} \frac{\arctan x}{x-1} \, dx &\stackrel{(x=1+\epsilon u)}{=} \int_{1}^{2+\epsilon} \frac{\arctan(1+\epsilon u)}{u} \, du \\ &\xrightarrow[\epsilon\downarrow0]{} \int_{1}^{2} \frac{\arctan(1)}{u} \, du = \frac{\pi}{4}\log 2. \end{align*}

$$\arctan(t^2)\ln(\frac{t^2-1}{t-1})\geq\int\limits_t^{t^2} \frac{\arctan(x)}{x-1}dx\geq\arctan(t)\ln(\frac{t^2-1}{t-1})$$ By squeeze theorem, the limit is $\frac{\pi}{4}\ln(2)$