Evaluate $\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$ 
I want to find a quick way of evaluating $$\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$$

This problem appeared on the qualifying round of MIT's 2014 Integration Bee, which leads me to think there should be a shortish way (no more than three minutes by hand) to solve it. Examining the indefinite integral (thanks to WolframAlpha) hasn't particularly helped me:
$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=-\frac{2}{3} \left(\sqrt{x^2+1}-2x\right) \sqrt{x+\sqrt{x^2+1}}+C$$
The bounds on the integral hint at trigonometric substitution, but I got nowhere by trying $x=\tan u$.
I also noticed that we can transform the integral by multiplying $\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}$ in the first square root, but there didn't seem to be anything to do after that either.

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int \sqrt{x+\sqrt{x^2+1}}dx$$
Now Let $$\left(x+\sqrt{x^2+1}\right)=t^2....................\color{red}\checkmark$$ 
Then $$\displaystyle \frac{\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}dx = 2tdt$$
So We get $$\displaystyle dx = \frac{2\sqrt{x^2+1}}{t}dt\;, $$ Now for Calculation of $$\sqrt{x^2+1}$$ in terms of $t\;,$ We are Using
$$\displaystyle \left(x+\sqrt{x^2+1}\right)\cdot \left(x-\sqrt{x^2+1}\right) = t^2\cdot \left(x-\sqrt{x^2+1}\right)$$
So We get $$\displaystyle \left(\sqrt{x^2+1}-x\right)=\frac{1}{t^2}..................\color{red}\checkmark$$
So We Get $$\displaystyle \sqrt{x^2+1}dx = \frac{t^4+1}{2t^2}dt$$
So We get $$\displaystyle I = \int\frac{t\cdot 2(t^4+1)}{2t^3}dt = \int \left(t^2+t^{-2}\right) dt = 2t-\frac{1}{t}+\mathcal{C}$$
So We Get $$\displaystyle \int\sqrt{x+\sqrt{x^2+1}}dx = 2\sqrt{x+\sqrt{x^2+1}}+\sqrt{x-\sqrt{x^2+1}}+\mathcal{C}$$
