evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
 \int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C 
\end{align*}$$
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:
$$ \begin{align*}
t  \arcsin\left(\frac{2t}{1+t^2}\right)&- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\
&= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt
\end{align*} $$
 A: I think you made a simplification error. We have
$$\begin{align*}
\frac{d}{dt}\arcsin\left(\frac{2t}{1+t^2}\right) &= \frac{1}{\sqrt{1 - \frac{4t^2}{(1+t^2)^2}}}\left(\frac{2t}{1+t^2}\right)'\\
&= \frac{(1+t^2)}{\sqrt{(1+t^2)^2-4t^2}}\left(\frac{2(1+t^2)-4t^2}{(1+t^2)^2}\right)\\
&= \frac{(1+t^2)}{\sqrt{(1-t^2)^2}}\left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\\
&= \frac{2(1-t^2)}{(1+t^2)\sqrt{(t^2-1)^2}}
=\frac{2(1-t^2)}{(1+t^2)|t^2-1|}.
\end{align*}$$
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.
A: The Weierstrass substitution in a slightly different form from that in which I'm accustomed to seeing it will do it.
We have
$$
\begin{align}
t & = \tan\frac x2 \\  \\
\frac{2\;dt}{1+t^2} & = dx \\  \\
\frac{2t}{1+t^2} & = \sin x \\  \\
\frac{1-t^2}{1+t^2} & = \cos x
\end{align}
$$
That's the usual Weierstrass substitution.  Now, differentiate the first line above to get
$$
dt = \frac12\sec^2\frac x2\;dx
$$
so this is
$$
\frac{dx}{2\cos^2\frac x2}
$$
and by the cosine half-angle formula, this is
$$
\frac{dx}{1+\cos x}.
$$
By the third line above, we have
$$
\arcsin\left(\frac{2t}{1+t^2}\right) = \arcsin \sin x = x
$$
(if $0\le x\le \pi/2$).  Therefore the desired integral becomes
$$
\int \frac{x\;dx}{1+\cos x} = \int x\;dt.
$$
Integrating by parts, we get
$$
xt - \int t\;dx = x\tan\frac x2 - \int \tan \frac x2 \; dx = x\tan\frac x2 - 2\log\cos\frac x2 + C.
$$
As $t$ goes from $0$ to $\sqrt{3}$, $x$ goes from $0$ to $\pi/3$, and there you have it.
Correction: As $t$ goes from $0$ to $\sqrt{3}$, the function $t\mapsto2t/(1+t^2)$ goes from $0$ up to $1$ and then starts going down again.  It reaches its maximum at $t=1$.  So $\sin x$ goes from $0$ up to $1$ and then starts going down again.  Thus $x$ goes from $0$ to $2\pi/3$.
This creates problems when one says $\arcsin\sin x = x$, since that applies when $x$ is between $0$ and $\pi/2$.  For $x$ between $\pi/2$ and $2\pi/3$, we'd have $\arcsin\sin x = \pi-x$ and we need to examine that interval separately.
A: 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)$
