Let, $\sqrt {x + 1} = t$
Hence, $\frac{1}{{2\sqrt {x + 1} }}dx = dt$
Thus, $dx = 2\sqrt {x + 1} dt$
The given integral = $\int\limits_{1/\sqrt 3 }^1 {\frac{{2\sqrt {x + 1} dt}}{{{{(x + 1)}^{1/2}} + {{(x + 1)}^{3/2}}}}} $
=$\int\limits_{1/\sqrt 3 }^1 {\frac{{2tdt}}{{t + {t^3}}}} $
=$2\int\limits_{1/\sqrt 3 }^1 {\frac{{dt}}{{1 + {t^2}}}} $
=$2\left[ {{{\tan }^{ - 1}}t|_{1/\sqrt 3 }^1} \right]$
=$2\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = \frac{\pi }{6}$
