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


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


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 $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$

  • $\begingroup$ See also : math.stackexchange.com/questions/589288/… and math.stackexchange.com/questions/322690/… $\endgroup$ – lab bhattacharjee Nov 28 '17 at 10:19
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    $\begingroup$ "1.I don't know what to with the $\pm$" Use that you know $x\geq 2$ to help you choose one of the solutions. $\endgroup$ – Arthur Nov 28 '17 at 10:21
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    $\begingroup$ "without a calculator..." First of all, a calculated value is not a mathematical solution. Second of all, maybe the calculator has an algebraic solver (like Macsyma). Is your real question "does the anti-derivative exist in closed form?" ? $\endgroup$ – Carl Witthoft Nov 28 '17 at 16:09
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    $\begingroup$ Note that your technique for simplifying the nested radicals only works if the expressions converge. When the expressions don't converge, it will still give you a value, but that value will be meaningless. Now, in this case, you can show that both sequences are increasing and bounded above, and so must converge. But in general, you should be very cautious with the "$y = f(x, y)$, so $y = f(x, f(x, f(x, ...)))$" concept. $\endgroup$ – Paul Sinclair Nov 28 '17 at 17:32
  • $\begingroup$ @CarlWitthoft I agree with you but I would have liked that better if it was an indefinite integral.;-) $\endgroup$ – tatan Nov 29 '17 at 3:25

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}

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  • $\begingroup$ Can you explain your substitution of t in the second integral? $\endgroup$ – tatan Nov 29 '17 at 3:18
  • $\begingroup$ @tatan - if $t = \sqrt{1+4x}$, then $x = \frac{t^2 + 1}4$, so $dx = \frac t2 dt$. $\endgroup$ – Paul Sinclair Nov 29 '17 at 3:37
  • $\begingroup$ @PaulSinclair Thanks $\endgroup$ – tatan Nov 29 '17 at 3:43
  • $\begingroup$ correction: $x = \frac{t^2 -1}4$. $\endgroup$ – Paul Sinclair Nov 29 '17 at 3:44



For real $y,$ $\sqrt{y^2}=|y|\ge0$

and $\sqrt{1+4x}\ge3$ for $2\le x\le12\implies1-\sqrt{1+4x}<0$


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

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$ \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

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For 2. notice that


Now the second integral is pretty easy


for the first one instead


let $2\sqrt{x}=\sinh y$, so $x=\frac{\sinh^2 y}{4}$ and $dx=\frac{\cosh y \sinh y}{2}\ dy$. Hence

$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{\frac{\cosh y \sinh y}{2}}{\frac{\sinh^2}{4} y\sqrt{1+\sinh^2 y}}dy=\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{dy}{\sinh y}=$$


Now let $e^y=z$, so $e^y\ dy=dz$


where $\eta=e^{\sinh^{-1}{2\sqrt2}}$ and $\psi={e^{\sinh^{-1}{4\sqrt3}}}.$




$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx==\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{12}}}-1}{e^{\sinh^{-1}{2\sqrt{12}}}+1}\right|-\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{2}}}-1}{e^{\sinh^{-1}{2\sqrt{2}}}+1}\right|+4.$$

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For any:


where $\frac{1\pm\sqrt{1+4x}}{2x}$ is defined for domain {a, b}

$$\frac{1}{2}\int_a^{b}\frac{1}{x}dx \pm \int_a^{b}\frac{\sqrt{1+4x}}{2x}dx$$ $$\frac{1}{2}(\ln{|b|} - \ln{|a|}) + \int_a^{b}\frac{\sqrt{1+4x}}{2x}dx - \int_a^{b}\frac{\sqrt{1+4x}}{2x}dx$$

$$\frac{1}{2}(\ln{|b|} - \ln{|a|})$$

Check if {2, 12} is a valid. I'm tired, so I won't picture what $\frac{1}{2x}\pm\frac{\sqrt{1+4x}}{2x}$ looks like. But you only have to check on the functions' decompositions.

Defined domains for each decompositions: $$\frac{1}{2x} \quad defined \quad \overline{\{0\}}$$


$$\frac{\sqrt{1+4x}}{2x} \Rightarrow \frac{1}{2x}\sqrt{1+4x}$$ $$1+4x<0 \Rightarrow x < -\frac{1}{4}$$ $$\frac{1}{2x}\sqrt{1+4x} \quad defined \quad \overline{\{0\} \; \cup \; x < -\frac{1}{4}}$$

therefore, $$\frac{1\pm\sqrt{1+4x}}{2x} \quad defined \quad \overline{\{0\} \; \cup \; x < -\frac{1}{4}}$$

so domain {2, 12} is safe and $\frac{1}{2}(\ln{|b|} - \ln{|a|})$ applies

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  • $\begingroup$ I think you should add $=$ signs in between $\endgroup$ – tatan Nov 28 '17 at 10:35
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    $\begingroup$ I did not understand how you got from the $\pm$ to separating the $+$ and $-$ $\endgroup$ – tatan Nov 28 '17 at 10:36
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    $\begingroup$ Huh? $a\pm b$ does not mean $a+b-b$. $\endgroup$ – Jack M Nov 28 '17 at 11:13
  • $\begingroup$ Yes it does. Coz it encompasses 2 functions, exactly opposite of each other $\pm b$, both $b$ and $-b$. Makes sense coz we have a square root function. Gives us 2 possibilities as value. So if we integrate, $b$ and $-b$ integrals cancel out. $\endgroup$ – Dehbop Nov 28 '17 at 11:18
  • $\begingroup$ OK, I MIGHT be wrong if $\frac{1\pm\sqrt{1+4x}}{2x}$ is the wrong identity. Honestly, I'm not that sure about that identity. Check with the asker. Though $\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}$, without sign, does seem to imply $\pm$ so it should carry out that there is $\pm$ in $\frac{1\pm\sqrt{1+4x}}{2x}$ $\endgroup$ – Dehbop Nov 28 '17 at 11:23

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