TITLE: Integral with 4 radicals-hat
QUESTION [8 upvotes]: I'd like to find out a simple way for calculating the value of:

$$\int_{0}^{1}\sqrt{1+\sqrt{1 + {\sqrt{1+ \sqrt{x}}}}}\,dx .$$ 

Of course, I thought of some variable change, but it seems pretty complicated. 
On the other hand, I wonder if there can be made a generalization when having 
to deal with the expression with $k$ radicals, $k>1$.

REPLY [4 votes]: Let
$$\begin{eqnarray*}
u &=&\sqrt{1+\sqrt{1+\sqrt{x}}} \Leftrightarrow &x=\left( \left( u^{2}-1\right) ^{2}-1\right)
^{2}=u^{8}-4u^{6}+4u^{4}.
\end{eqnarray*}$$
Since $$\begin{equation*}
dx=\left( 8u^{7}-24u^{5}+16u^{3}\right) du
\end{equation*}$$ we have 
$$I :=\int_{0}^{1}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}dx\\=\int_{\sqrt{2}}^{\sqrt{1+\sqrt{2}}}\sqrt{1+u}\left(8u^{7}-24u^{5}+16u^{3}\right) du.\quad\textit{(computation below)}^†
$$ Each term can be integrated using the substitution $t=\sqrt{1+u}$ $$\begin{equation*}
\int_{a}^{b}\sqrt{1+u}u^{n}du=2\int_{\sqrt{1+a}}^{\sqrt{1+b}}t^{2}\left(
t^{2}-1\right) ^{n}\,dt,\quad a=\sqrt{2},b=\sqrt{1+\sqrt{2}}.
\end{equation*}$$
Generalization to $k=5$ radicals $$\begin{equation*}
J:=\int_{0}^{1}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}}dx.
\end{equation*}$$ Similarly to above the substitution is now
$$\begin{eqnarray*} v &=&\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}\Leftrightarrow  x=\left( \left( \left( v^{2}-1\right) ^{2}-1\right) ^{2}-1\right) ^{2} \\  x &=& v^{16}-8v^{14}+24v^{12}-32v^{10}+14v^{8}+8v^{6}-8v^{4}-1,
\end{eqnarray*}$$ and
$$\begin{equation*} dx=\left(
16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv.
\end{equation*}$$ 

Hence
$$\begin{eqnarray*}
J &=&\int_{\alpha }^{\beta }\sqrt{1+v}\left(
16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv \\
\alpha  &=&\sqrt{1+\sqrt{2}},\beta =\sqrt{1+\sqrt{1+\sqrt{2}}}.
\end{eqnarray*}$$
--
†In SWP I obtained
$$\begin{eqnarray*}
I &=&-\frac{26\,704}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{\sqrt{2}+1}
\sqrt{2} \\&&+\frac{83\,584}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{\sqrt{2}+1} \\
&&+\frac{344\,096}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}} \\
&&+\frac{67\,328}{109\,395}\sqrt{\sqrt{2}+1} \\
&&-\frac{256}{3003}\sqrt{\sqrt{2}+1}\sqrt{2} \\
&&-\frac{17\,168}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{2} \\
&\approx &1.584\,9.
\end{eqnarray*}$$