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.