Substitue $e^{x}=t$ and $e^{x} d x=d t$
Therefore
$I=\int \frac{\cot ^{-1}(t)}{t^{2}} d t$
Integrating it by part, taking $\cot ^{-1}(t)$ as first function and $t^{2}$ as second function, we get $I=-\frac{\cot ^{-1}(t)}{t}-\int \frac{1}{t\left(t^{2}+1\right)} d t$
$\int \frac{1}{t\left(t^{2}+1\right)} d t=P$
$t^{2}=u, \quad 2 t d t=d u$
$P=\frac{1}{2} \int \frac{1}{u(u+1)} d u$
$=\frac{1}{2} \int\left(\frac{1}{u}-\frac{1}{(u+1)}\right) d u$
$=\frac{1}{2}(\log u-\log u+1)+C$
$I=-\frac{\cot ^{-1}(t)}{t}-\frac{1}{2}(\log |u|-\log |u+1|)+C$
$=\frac{1}{2}\left(\log \left|t^{2}+1\right|-\log \left|t^{2}\right|-\frac{2 \cot ^{-1}(t)}{t}\right)+C$
$=\frac{1}{2}\left(\log \left|e^{2 x}+1\right|-\log \left|e^{2 x}\right|-\frac{2 \cot ^{-1}\left(e^{x}\right)}{e^{x}}\right)+C$
$=\frac{1}{2} \log \left|e^{2 x}+1\right|-\frac{\cot ^{-1}\left(e^{x}\right)}{e^{x}}-x+C$
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$\sin ^{-1}\left(\sum_{i=1}^{\infty} x^{i+1}-x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^i\right)=\frac{\pi}{2}-\cos ^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^i-\sum_{i=1}^{\infty}(-x)^i\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is. . . . .
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[0, \pi]$, respectively.)
$f(x) = \sqrt {\frac{{4 - {x^2}}}{{\left[ x \right] + 2}}} $ is $($ where $[.] \rightarrow G.I.F.)$