We have the identity
[tex]\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}[/tex]
Take the square root of both sides and rearrange terms on the right to get
[tex]\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}[/tex]
Decrementing n gives
[tex]\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}[/tex]
and substituting the previous expression into this, we have
[tex]\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }[/tex]
Continuing in this fashion, after k steps we would have
[tex]\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}[/tex]
After a total of n - 2 steps, we arrive at
[tex]\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}[/tex]
Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to
[tex]\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}[/tex]
