Can this definite integral involving series be solved without a calculator? I got this question today but I can't see if there is any good way to solve it by hand.

Evaluate the definite integral
$$\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}\,\mathrm{d}x$$
where the series in the numerator and denominator continue infinitely.

If you let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$, solving for $y$ we get $y=\frac{1\pm\sqrt{1+4x}}{2}$. And similarly for the denominator we have $z=\sqrt{xz}$. So $z=x$. So the integral simplifies to
$$\int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$
Now my problems are

*

*I don't know what to with the $\pm$.


*I tried to solve the integral by separating it as a sum of two fractions. But I can't solve $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$

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}
