# Evaluating $$\displaystyle{\int_{{0}}^{{\sqrt{{{3}}}}}}{\arcsin{{\left({\frac{{{2}{t}}}{{{1}+{t}^{{2}}}}}\right)}}}{\left.{d}{t}\right.}$$

Evaluating ${\int }_{0}^{\sqrt{3}}\mathrm{arcsin}\left(\frac{2t}{1+{t}^{2}}\right)dt$
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Jambrichp2w2
We have
$\frac{d}{dt}\mathrm{arcsin}\left(\frac{2t}{1+{t}^{2}}\right)=\frac{1}{\sqrt{1-\frac{4{t}^{2}}{{\left(1+{t}^{2}\right)}^{2}}}}{\left(\frac{2t}{1+{t}^{2}}\right)}^{\prime }$
$=\frac{\left(1+{t}^{2}\right)}{\sqrt{{\left(1+{t}^{2}\right)}^{2}-4{t}^{2}}}\left(\frac{2\left(1+{t}^{2}\right)-4{t}^{2}}{{\left(1+{t}^{2}\right)}^{2}}\right)$
$=\frac{\left(1+{t}^{2}\right)}{\sqrt{{\left(1-{t}^{2}\right)}^{2}}}\left(\frac{2\left(1-{t}^{2}\right)}{{\left(1+{t}^{2}\right)}^{2}}\right)$
$=\frac{2\left(1-{t}^{2}\right)}{\left(1+{t}^{2}\right)\sqrt{{\left({t}^{2}-1\right)}^{2}}}$
$=\frac{2\left(1-{t}^{2}\right)}{\left(1+{t}^{2}\right)|{t}^{2}-1|}$