<s>[INST] Another series involving $\log (3)$ I will show that

$$\sum_{n = 0}^\infty \left (\frac{1}{6n + 1} + \frac{1}{6n + 3} + \frac{1}{6n + 5} - \frac{1}{2n + 1} \right ) = \frac{1}{2} \log (3).$$

My question is can this result be shown more simply then the approach given below? Perhaps using Riemann sums?
Denote the series by $S$ and let $S_n$ be its $n$th partial sum.
\begin{align}
S_n &= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 3} + \frac{1}{6k + 5} - \frac{1}{2k + 1} \right )\\
&= \sum_{k = 0}^n \left (\frac{1}{6k + 1} + \frac{1}{6k + 2} + \frac{1}{6k + 3} + \frac{1}{6k + 4} + \frac{1}{6k + 5} + \frac{1}{6k + 6} \right )\\
& \quad - \sum_{k = 0}^n \left (\frac{1}{2k + 1} + \frac{1}{2k + 2} \right ) - \frac{1}{2} \sum_{k = 0}^n \left (\frac{1}{3k + 1} + \frac{1}{3k + 2} + \frac{1}{3k + 3} \right )\\
& \qquad + \frac{1}{2} \sum_{k = 0}^n \frac{1}{k + 1}\\
&= H_{6n + 3} - H_{2n + 2} - \frac{1}{2} H_{3n + 3} + \frac{1}{2} H_{n + 1}.
\end{align}
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n \frac{1}{k}$. Since $H_n = \log (n) + \gamma + o(1)$ where $\gamma$ is the Euler-Mascheroni constant we see that
$$S_n = \log (6n) - \log (2n) - \frac{1}{2} \log (3n) + \frac{1}{2} \log (n) + o(1) = \frac{1}{2} \log (3) + o(1).$$
Thus
$$S = \lim_{n \to \infty} S_n = \frac{1}{2} \log (3).$$
 [/INST] Your sum is$$\sum_{n\ge0}\int_0^1x^{6n}(1-2x^2+x^4)dx=\int_0^1\dfrac{(1-x^2)^2}{1-x^6}dx=\int_0^1\dfrac{1-x^2}{1+x^2+x^4}dx,$$where the first $=$ uses monotone convergence. Since$$1+x^2+x^4=(1+x^2)^2-x^2=\prod_\pm(1\pm x+x^2),$$you can show as an exercise that this integral is$$\left[\dfrac12\ln\dfrac{1+x+x^2}{1-x+x^2}\right]_0^1=\dfrac12\ln3.$$
</s>