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

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

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

Solution 1:

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}

Solution 2:

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

Solution 3:

$ \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