Evaluate $\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$

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.

$\bf{My\; Solution::}$ 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 $$\displaystyle \int\sqrt{x+\sqrt{x^2+1}}dx = 2\sqrt{x+\sqrt{x^2+1}}+\sqrt{x-\sqrt{x^2+1}}+\mathcal{C}$$

$$ \begin{align} \int_0^{1/\sqrt3}\sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x &=\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\mathrm{d}\tan(\theta)\tag{1}\\ &=\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{2}\\ &=\frac1{\sqrt[4]3}-\int_0^{\pi/6}\tan(\theta)\frac{\sec^2(\theta)+\tan(\theta)\sec(\theta)}{2\sqrt{\tan(\theta)+\sec(\theta)}}\,\mathrm{d}\theta\tag{3}\\ &=\frac1{\sqrt[4]3}-\frac12\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\mathrm{d}\sec(\theta)\tag{4}\\ &=\frac12+\frac12\int_0^{\pi/6}\sec(\theta)\frac{\sec^2(\theta)+\tan(\theta)\sec(\theta)}{2\sqrt{\tan(\theta)+\sec(\theta)}}\,\mathrm{d}\theta\tag{5}\\ &=\frac12+\frac14\int_0^{\pi/6}\sqrt{\tan(\theta)+\sec(\theta)}\,\sec^2(\theta)\,\mathrm{d}\theta\tag{6}\\[6pt] &=\frac23\tag{7} \end{align} $$ 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 $(2)$