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.
 A: $\bf{My\; Solution::}$ 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 $$\displaystyle \int\sqrt{x+\sqrt{x^2+1}}dx = 2\sqrt{x+\sqrt{x^2+1}}+\sqrt{x-\sqrt{x^2+1}}+\mathcal{C}$$
A: $$
\begin{align}
\int_0^{1/\sqrt3}\sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x
&=\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\mathrm{d}\tan(\theta)\tag{1}\\
&=\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{2}\\
&=\frac1{\sqrt[4]3}-\int_0^{\pi/6}\tan(\theta)\frac{\sec^2(\theta)+\tan(\theta)\sec(\theta)}{2\sqrt{\tan(\theta)+\sec(\theta)}}\,\mathrm{d}\theta\tag{3}\\
&=\frac1{\sqrt[4]3}-\frac12\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\mathrm{d}\sec(\theta)\tag{4}\\
&=\frac12+\frac12\int_0^{\pi/6}\sec(\theta)\frac{\sec^2(\theta)+\tan(\theta)\sec(\theta)}{2\sqrt{\tan(\theta)+\sec(\theta)}}\,\mathrm{d}\theta\tag{5}\\
&=\frac12+\frac14\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{6}\\[6pt]
&=\frac23\tag{7}
\end{align}
$$
Explanation:
$(1)$: substitution $x\mapsto\tan(\theta)$
$(2)$: $\frac{\mathrm{d}}{\mathrm{d}\theta}\tan(\theta)=\sec^2(\theta)$
$(3)$: integrate by parts
$(4)$: $\frac{\mathrm{d}}{\mathrm{d}\theta}\sec(\theta)=\tan(\theta)\sec(\theta)$
$(5)$: integrate by parts
$(6)$: simplification
$(7)$: $\frac43$ of $(6)$ minus $\frac13$ of $(2)$
A: If you have dealt with the explicit formulas for arcsinh and arccosh you might recognize that
$${\rm arcsinh}\,\,  x = \log (x + \sqrt{x^2 + 1})$$
So you are being asked to integrate
$$\int_0^{1 \over \sqrt{3}} e^{{1 \over 2} {\rm arcsinh}\,x}\,dx$$
So you are led to substituting $x = \sinh y$ and the integral is
$$\int_0^{\sinh^{-1}{1 \over \sqrt{3}}} e^{y \over 2} \cosh y \,dy$$
$$= \int_0^{\sinh^{-1} {1 \over \sqrt{3}}} {e^{3y \over 2} + e^{-{y \over 2}} \over 2} \,dy$$
This is now an easy calculus problem although the result might be annoying due to the bounds of integration here. However, $\sinh^{-1} {1 \over \sqrt{3}} = \ln ({1 \over \sqrt{3}}+ \sqrt{{1 \over 3} + 1})= \ln \sqrt{3} = {1 \over 2}\ln 3$, so the result is
$$\big({1 \over 3} e^{3y \over 2} - e^{-{y \over 2}}\big)\bigg\vert_0^{{1 \over 2} \ln 3}$$
$$= {1 \over 3} e^{{3 \over 4}\ln 3} - e^{-{\ln 3 \over 4}} + {2 \over 3}$$
$$= {1 \over 3} 3^{3 \over 4} - 3^{-{1 \over 4}} + {2 \over 3}$$
$$ = {2 \over 3}$$
A: Also works well with $x=\frac12\left(t-\dfrac1t\right)$.
$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=\frac12\int\sqrt{\frac12\left(t-\frac1t\right)+\frac12\left(t+\frac1t\right)}\left(1+\frac1{t^2}\right)\,dt\\
=\frac12\int\left(t^{1/2}+t^{-3/2}\right)\,dt=\frac13t^{3/2}-t^{-1/2}.$$
To transform the bounds, $t^2-2xt-1=0$ i.e. $t=x+\sqrt{x^2+1}$, giving $1$ to $\sqrt3$.
A: Hint
I suppose that the change of variable is $x=\sinh(y)$, $dx=\cosh(y)\,dy$ and so $$I=\int\sqrt{x+\sqrt{x^2+1}}\,dx=\int\sqrt{\sinh(y)+\cosh(y)} \cosh(y)\,dy=\int \cosh(y)\,e^{y/2}\,dy$$ $$I=\int\frac{e^y+e^{-y}}2\,e^{y/2}\,dy=\frac{1}{3} e^{3 y/2}-e^{-y/2}$$ and, back to $x$, the formula given by Wolfram Alpha.
A: Why not just substitute the whole complicated integrand?
This seems to be the simplest approach, straighforward, no guesswork necessary.
$$u = \sqrt{\sqrt{x^2+1}+x}$$
Then we have
$$u(x=0) = 1, u(x=\frac{1}{\sqrt{3}})= \sqrt[4]{3}\tag{*}$$
$$x = \frac{u^4-1}{2 u^2}$$
$$\,dx = (\frac{1}{u^3}+u)\,du$$
And the indefinite integral is elementary
$$i = \int \left(\frac{1}{u^3}+u\right) u \, du = \frac{u^3}{3}-\frac{1}{u}$$
In the limits (*) it becomes $\frac{2}{3}$.
A: Using formula:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$
so:
$$\sqrt{x+\sqrt{x^2+1}}=\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}$$
and then:
$$\int \left(\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}\right) \, dx=\frac{1}{3} \sqrt{2} (-i+x)^{3/2}+\frac{1}{3} \sqrt{2} (i+x)^{3/2}+C$$
