Limit of $\lim_{t\to 1^+} \int_t^{t^2} \frac{\arctan(x)}{x-1} dx$ I need to evaluate $$\lim_{t\to 1^+} \int_t^{t^2} \frac{\arctan(x)}{x-1} \; dx.$$
I have tried the partial integrating, and I got confused. Could someone help me, please?
 A: 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*}
A: $$\arctan(t^2)\ln(\frac{t^2-1}{t-1})\geq\int\limits_t^{t^2} \frac{\arctan(x)}{x-1}dx\geq\arctan(t)\ln(\frac{t^2-1}{t-1})$$
By squeeze theorem, the limit is $\frac{\pi}{4}\ln(2)$
A: $artan (x) \to \frac {\pi} 4$ as $ x \to 1$. Hence the lmit  is $ \frac {\pi} 4 \lim \int_t^{t^{2}} \frac 1 {x-1} \, dt=(log \, 2 ) \frac {\pi} 4  $
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\lim_{t \to 1^{\large +}}\int_{t}^{t^{2}}
{\arctan\pars{x} \over x - 1}\,\dd x}
\\[5mm] = &\
\lim_{t \to 1^{\large +}}\int_{t}^{t^{2}}
{\arctan\pars{x} - \arctan\pars{1} \over x - 1}\,\dd x\ +\
\lim_{t \to 1^{\large +}}\underbrace{\int_{t}^{t^{2}}
{\arctan\pars{1} \over x - 1}\,\dd x}
_{\ds{=\ {\pi \over 4}\,\ln\pars{t + 1}}}\label{1}\tag{1}
\end{align}

Note that
  $\ds{\left.{\arctan\pars{x} - \arctan\pars{1} \over x - 1}
\,\right\vert_{\ 1\ <\ t\ <\ x\ <\ t^{2}} =
\left.{1 \over \xi^{2} + 1}
\,\right\vert_{\ 1 <\ \xi\ <\ x}}$
$\ds{\implies
\bbx{{1 \over x^{2} + 1} <
\left.{\arctan\pars{x} - \arctan\pars{1} \over x - 1}
\,\right\vert_{\ 1\ <\ t\ <\ x\ <\ t^{2}} < {1 \over 2}}}$

The first limit in \eqref{1} vanishes out because
$$
\arctan\pars{t^{2}} - \arctan\pars{t} <
\int_{t}^{t^{2}}
{\arctan\pars{x} - \arctan\pars{1} \over x - 1}\,\dd x <
{t\pars{t - 1} \over 2}
$$
such that \eqref{1} becomes
$$
\bbox[10px,#ffd]{\lim_{t \to 1^{\large +}}\int_{t}^{t^{2}}
{\arctan\pars{x} \over x - 1}\,\dd x} =
\bbx{{\pi \over 4}\,\ln\pars{2}}
$$
