<s>[INST] 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 $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$
 [/INST] 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
$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx==\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{12}}}-1}{e^{\sinh^{-1}{2\sqrt{12}}}+1}\right|-\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{2}}}-1}{e^{\sinh^{-1}{2\sqrt{2}}}+1}\right|+4.$$
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