evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
 \int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&amp;=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&amp;=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C 
\end{align*}$$
So
$$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^2}\right)=\pi/\sqrt3+2\ln2.$$

However the result seems to be $ \pi/\sqrt3 $ only. 
Why is there this $ 2\ln2 $?

Detail:

$$ \begin{align*}
t  \arcsin\left(\frac{2t}{1+t^2}\right)&amp;- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\
&amp;= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\
&amp;=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt
\end{align*} $$
I think you made a simplification error. We have
$$\begin{align*}
\frac{d}{dt}\arcsin\left(\frac{2t}{1+t^2}\right) &amp;= \frac{1}{\sqrt{1 - \frac{4t^2}{(1+t^2)^2}}}\left(\frac{2t}{1+t^2}\right)&#39;\\
&amp;= \frac{(1+t^2)}{\sqrt{(1+t^2)^2-4t^2}}\left(\frac{2(1+t^2)-4t^2}{(1+t^2)^2}\right)\\
&amp;= \frac{(1+t^2)}{\sqrt{(1-t^2)^2}}\left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\\
&amp;= \frac{2(1-t^2)}{(1+t^2)\sqrt{(t^2-1)^2}}
=\frac{2(1-t^2)}{(1+t^2)|t^2-1|}.
\end{align*}$$
You then cancelled to get
$$-\frac{2}{1+t^2}.$$
However, that cancellation is only valid if $t^2-1\geq 0$, i.e., if $|t|\geq 1$. Yet your integral covers a region that includes places where you get $t^2\lt 1$, so that the cancellation is not valid over the entire interval. Try doing it by splitting the integral as an integral over $[0,1]$ and over $[1,\sqrt{3}]$, being careful with the signs.
part of integral solution is $\ln(1+t^2)$. When you insert integral bounds you get $\ln(1+(\sqrt{3})^2)-\ln(1+(0)^2)$$=\ln(4)-ln(1)$$=2\ln(2)$