Can this definite integral involving series be solved without a calculator?

As you did, let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$. Clearly $y\ge0$ and $y$ satisfies $$ y^2-y-x=0 $$ from which you have $$ y=\frac{1\pm\sqrt{4x+1}}{2}. $$ Since $x\in[2,12]$ and $y\ge0$, you must choose "$+$". Since if you choose "$-$", then $$ y=\frac{1-\sqrt{4x+1}}{2}<0. $$ Now, under $t=\sqrt{1+4x}$ \begin{eqnarray} &&\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}dx\\ &=&\int_2^{12}\frac{1+\sqrt{1+4x}}{2x}dx\\ &=&\int_2^{12}\frac{1}{2x}dx+\int_2^{12}\frac{\sqrt{1+4x}}{2x}dx\\ &=&\frac12\ln x\bigg|_2^{12}+\int_3^7\frac{t^2}{t^2-1}dt\\ &=&\frac12\ln6+\bigg(t+\frac12\ln\frac{t-1}{t+1}\bigg)\bigg|_3^7\\ &=&\frac12\ln6+4+\frac12\ln\frac32\\ &=&\ln3+4. \end{eqnarray}

Hint:

 1.

For real $y,$ $\sqrt{y^2}=|y|\ge0$

and $\sqrt{1+4x}\ge3$ for $2\le x\le12\implies1-\sqrt{1+4x}<0$

2.

Set $\sqrt{1+4x}=u\implies4x=u^2-1$