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 $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$
 A: 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 $\sqrt{1+4x}=u\implies4x=u^2-1$
A: 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}
A: $ \int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}dx$
Now, notice that the limits are defined for positive values of x. Also, 
$y=\frac{1\pm\sqrt{1+4x}}{2}<0$ if you consider the negative sign for $x>0$. 
But, $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}>0$ since the square root of a number is always positive for it to be a function (otherwise, for only one value of $x$, there'll be two values of y i.e., multiple mapping into range for only one value in the domain).
So, $y>0$. 
So, $I= \int_2^{12}\frac{1+\sqrt{1+4x}}{2x}dx$
Now, just substitute $4x+1=u^2$ (as mentioned in one of the above answers).
$\implies I=\int \frac{u}{u-1}du$ with appropriate limits
A: For 2. notice that
$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx=\int_{2}^{12}\frac{1+4x}{2x\sqrt{1+4x}}=\frac{1}{2}\left(\int_{2}^{12}\frac{1}{x\sqrt{1+4x}}dx+4\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}\right).$$
Now the second integral is pretty easy
$$2\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}=\left[\sqrt{1+4x}\right]_{2}^{12}=4,$$
for the first one instead
$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{2}^{12}\frac{dx}{x\sqrt{1+\left(2\sqrt{x}\right)^2}},$$
let $2\sqrt{x}=\sinh y$, so $x=\frac{\sinh^2 y}{4}$ and $dx=\frac{\cosh y \sinh y}{2}\ dy$. Hence
$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{\frac{\cosh y \sinh y}{2}}{\frac{\sinh^2}{4} y\sqrt{1+\sinh^2 y}}dy=\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{dy}{\sinh y}=$$
$$=2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy.$$
Now let $e^y=z$, so $e^y\ dy=dz$
$$2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy=2\int_{\eta}^{\psi}\frac{dz}{z^2-1},$$
where $\eta=e^{\sinh^{-1}{2\sqrt2}}$ and $\psi={e^{\sinh^{-1}{4\sqrt3}}}.$
$$2\int_{\eta}^{\psi}\frac{dz}{z^2-1}=\ln\left|\frac{z-1}{z+1}\right|_{\eta}^{\psi}=\ln\left|\frac{e^y-1}{e^y+1}\right|_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}=$$
$$=\ln\left|\frac{e^{\sinh^{-1}{2\sqrt{x}}}-1}{e^{\sinh^{-1}{2\sqrt{x}}}+1}\right|_{2}^{12},$$
so
$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx==\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{12}}}-1}{e^{\sinh^{-1}{2\sqrt{12}}}+1}\right|-\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{2}}}-1}{e^{\sinh^{-1}{2\sqrt{2}}}+1}\right|+4.$$
