How to obtain the sum of the series $\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}$? How to prove what follows?

$$\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\frac{2^{\frac{1}{3}}}{3}\ln\left(\frac{\sqrt{2^{\frac{2}{3}}+2^{\frac{1}{3}}+1}}{2^{\frac{1}{3}}-1}\right)+\frac{\sqrt[3]{2}}{3}\arctan\left(\frac{2^{\frac{2}{3}}+1}{\sqrt{3}}\right)-\frac{2^{\frac{1}{3}}\pi}{6\sqrt{3}}$$

My attempt:
$$\sum_{n=0}^{\infty}\frac{1}{2^n(3n+1)}=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}|_{x=1}$$
We put $$S(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}\implies S^{'}(x)=\sum_{n=0}^{\infty}\frac{x^{3n}}{2^n(3n+1)}$$
$$S^{'}(x)=\sum_{n=0}^{\infty}\frac{(\frac{x^3}{2})^n}{3n+1}=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(1-\frac{3n}{3n+1})=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n\frac{3n}{3n+1}=\frac{1}{2-\frac{x^3}{2}}-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})=\alpha-\beta$$
Where
$$\beta=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})$$
So $$\beta=?$$
Waiting for your help to find a beta or prove equal above.
 A: Observe that
$$f(x):=\sum_{n=0}^\infty\frac{x^n}{3n+1}\implies f'(x)=\sum_{n=0}^\infty\frac{nx^{n-1}}{3n+1}$$ and
$$3xf'(x)+f(x)=\sum_{n=0}^\infty\frac{3n+1}{3n+1}x^n=\frac1{1-x}.$$
The solution of the homogeneous part of this linear ODE is
$$f_h(x)=\frac c{\sqrt[3]x},$$
and by variation of the constant
$$f(x)=\frac1{\sqrt[3]x}\left(\int\frac{\sqrt[3]x}{1-x}dx+c\right).$$
A: For all $|x|<1$ we have the elementary geometric series $$\sum_{n\geq 0}x^n=\frac{1}{1-x} \\\underbrace{=}^{x\to x/2}\sum_{n\geq 0}\frac{x^n}{2^n} =\frac{2}{2-x}$$ now replacing $x$ by $x^3$ and then on integration from $0$ to $1$ we have $$\sum_{n\geq 0}\frac{1}{2^n(3n+1)}=\int_0^1\frac{2}{2-x^3}dx$$ since $x^3-2=(x-\sqrt[3]2)(x^2-\sqrt[3]{2}x+\sqrt[3]{8})$ by partial fraction of the integrand we write the  last expression as $$-\int_0^1\frac{2}{x^3-2} dx=-\frac{2}{3\sqrt[3]4}\int_0^1\left(\color{red}{\frac{1}{x-\sqrt[3]{2}}}-\color{blue}{\frac{x+\sqrt[3]{16}}{x^2+\sqrt[3]{2}x +\sqrt[3]4}}\right)dx$$ It's is easy to see  that red integral $$\int_0^1\color{red}{\frac{1}{x-\sqrt[3]2}}dx =\ln(|x- \sqrt[3]2|)\bigg|_0^1=\ln(|1-\sqrt[3]2|)-\ln\sqrt[3]2\cdots(1)$$
Further note that $$\int_0^1\color{blue}{\frac{x+\sqrt[3]{16}}{x^2+\sqrt[3]{2}x+\sqrt[3]{4}}}dx=\frac{1}{2}\int_0^1\left(\frac{2x+\sqrt[3]2}{x^2+\sqrt[3]2 x+\sqrt[3]4}+\frac{3\cdot 
\sqrt[3]{2}}{x^2+\sqrt[3]2x+\sqrt[4]{4}}\right)dx$$ The last two integral are standard and elementary logarithm and arctangent integrals and integrating them we have $$\frac{1}{2}\ln(x^2+\sqrt[3]2x+\sqrt[3]4)+\sqrt[3]{2}\tan^{-1}\left(\frac{\sqrt[3]{4}x+1}{\sqrt 3}\right)\bigg|_0^1=\frac{\ln(1+\sqrt[3]{2}+\sqrt[3]{4})-\ln(\sqrt[3]{4})}{2}+\sqrt{3}\tan^{-1}\left(\frac{\sqrt[3]{4}+1}{\sqrt{3}}\right)-\frac{\pi}{2\sqrt{3}}\cdots(2)$$ subtract $(1)$ from $(2)$ and multiply by the factor $-\frac{2}{3\sqrt[3]{2}}$ and simplification gives us the desired result of the series.

$$\frac{2}{3\sqrt[3]{4}}\left(\sqrt{3}\tan^{-1}\left(\frac{\sqrt[3]{4}+1}{\sqrt 3}\right)+\frac{1}{2}\ln(1+\sqrt[3]{2}+\sqrt[3]{4})-\ln|\sqrt[3]{2}-1| -\frac{\pi}{2\sqrt{3}}\right)\approx 1.18143\cdots$$

A: You can modify your attempt a little bit to get an easier way of solving this problem:
$$\sum_{n=0}^{\infty} \dfrac{1}{2^n(3n+1)} = \sum_{n=0}^{\infty} \dfrac{1}{(\sqrt[3]{2})^{3n}(3n+1)} = \sqrt[3]{2} \sum_{n=0}^{\infty} \dfrac{1}{(\sqrt[3]{2})^{3n+1}(3n+1)}$$
And now you can find the value of $$S(x) = \sum_{n=0}^{\infty} \dfrac{x^{3n+1}}{3n+1}$$ and replace $x = \dfrac{1}{\sqrt[3]{2}}$ back into $S(x)$
A: Hint. One may prove that
$$
\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\sum_{n=0}^{\infty}\int_0^1\frac{x^{3n}}{2^{n}}dx=\int_0^1\sum_{n=0}^{\infty}\left(\frac{x^3}{2}\right)^{n}dx=\int_0^1\frac{2}{2-x^3}\:dx
$$
Hope you can take it from here.
