On the integral $\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}$ and the plastic constant

Question

We have,

$$\phi=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$$

$$P=\sqrt[3]{1+\sqrt[3]{1+\sqrt[3]{1+\cdots}}}$$

with golden ratio $\phi$ and plastic constant $P$. If,

$$\int_0^1 \frac{dx}{\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=\int_0^1 \frac{2}{1+\sqrt{1+4x}}dx=\frac{2}{\phi}-\ln \phi$$

given in this answer, would it follow that the integral,

$$\int_0^1 \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}=\,?$$

is expressible in terms of the plastic constant?

Answer 1

HINT:

Let $\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}=y\implies x+y=y^3\iff x=y^3-y$

$dx=3y^2-1$

$x=1\implies y\approx 1.3247179572447458$

$$\int \frac{dx}{\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots}}}}=\int\dfrac{3y^2-1}y\ dy=\dfrac{3y^2}2-\ln|y|+K$$