<s>[INST] If the value of integral in the image below is π then what is the value of y? I could not simplify
$$
\int_0^1 \sqrt{-1 + \sqrt{\frac{1+y}{x} - y}}\ dx
$$
I tried integration it in an online integrator but trust me the result is seriously daunting to be back traced to $\pi$ as a value, thereby determining $y$.  So I am looking for a rather clever trick to get through this one. 
 [/INST] Define 
$$f \colon [-1,\infty) \to [0,\infty), \, f(y) = \int \limits_0^1 \sqrt{\sqrt{\frac{1+y}{x} - y}-1} \, \mathrm{d} x \, .$$
Obviously, $f(-1) = 0$. For $y > -1$ we have
\begin{align}
f(y) &= \int \limits_0^1 \sqrt{\sqrt{\frac{1+y}{x} - y}-1} \, \mathrm{d} x \stackrel{\frac{1+y}{x} - y = \frac{1}{t^2}}{=} (1+y) \int \limits_0^1 \frac{2t}{(1+y t^2)^2} \sqrt{\frac{1-t}{t}} \, \mathrm{d} t \\
&\!\stackrel{\text{IBP}}{=} (1+y) \int \limits_0^1 \frac{t^2}{1+y t^2} \frac{1}{2\sqrt{t^3(1-t)}} \, \mathrm{d} t = \frac{1+y}{2} \int \limits_0^1 \frac{\sqrt{\frac{t}{1-t}}}{1+yt^2} \, \mathrm{d} t \\
&\!\!\!\!\stackrel{t = \frac{u}{1+u}}{=} \frac{1+y}{2} \int \limits_0^\infty \frac{\sqrt{u}}{1 + 2u + (1+y) u^2} \, \mathrm{d} u \stackrel{u = \frac{v^2}{\sqrt{1+y}}}{=} (1+y)^{1/4} \int \limits_0^\infty \frac{v^2}{1 + \frac{2}{\sqrt{1+y}} v^2 + v^4} \, \mathrm{d} v \, .
\end{align}
The remaining integral can be computed using the residue theorem or the Cauchy-Schlömilch substitution discussed here. The second method is faster and yields
\begin{align}
f(y) &= \frac{(1+y)^{1/4}}{2} \int \limits_{-\infty}^\infty \frac{\mathrm{d} v}{\left(v - \frac{1}{v}\right)^2 + 2\left(1 + \frac{1}{\sqrt{1+y}}\right)} \stackrel{\text{CS}}{=} \frac{(1+y)^{1/4}}{2} \int \limits_{-\infty}^\infty \frac{\mathrm{d} w}{w^2 + 2\left(1 + \frac{1}{\sqrt{1+y}}\right)} \\
&= \frac{(1+y)^{1/4}}{2} \frac{\pi}{\sqrt{2\left(1 + \frac{1}{\sqrt{1+y}}\right)}} = \frac{\pi}{2} \sqrt{\frac{1+y}{2(1+\sqrt{1+y})}} \, .
\end{align}
As can be seen from the original definition, $f$ is strictly increasing from $0$ to $\infty$, so the inverse function $f^{-1} \colon [0,\infty) \to [-1,\infty)$ exists. For $x \geq 0$ we have
$$ x = \frac{\pi}{2} \sqrt{\frac{1+f^{-1}(x)}{2(1+\sqrt{1+f^{-1}(x)})}} \, ,$$
which can be rewritten as a simple quadratic equation for $\sqrt{1+f^{-1}(x)}$. Its solution is
$$ f^{-1} (x) = \left(\frac{2 x}{\pi}\right)^2 \left[\frac{2x}{\pi} + \sqrt{2 + \left(\frac{2x}{\pi}\right)^2}\right]^2 - 1 \stackrel{x > 0}{=} \left(\frac{2 x}{\pi}\right)^4 \left[1 + \sqrt{1 + 2 \left(\frac{\pi}{2x}\right)^2}\right]^2 - 1 \, ,$$
so
$$ f^{-1}(\pi) = 16 \left(1 + \sqrt{\frac{3}{2}}\right)^2 - 1 = 39 + 16 \sqrt{6} \, . $$
</s>