Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$

Positive integrals $$\int_{0}^{1}\frac{2x(1-x)^2}{1+x^2}dx=\pi-3$$ and $$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi $$ (https://math.stackexchange.com/a/1618454/134791)

prove that $$3<\pi<\frac{22}{7}$$

Is there a similar argument for the following $\log (2)$ inequality? $$\frac{2}{3}<\log(2)<\frac{7}{10}$$

There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$. $$ \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx &=\frac{7}{10}-\log\left(2\right) \\ \end{align} $$

Therefore, $$-\int_0^1\frac{x^2(1-x)^2}{1+x^2}dx<0<\int_0^1\frac{x^4(1-x)^2}{1+x^2}dx$$

$$\frac{2}{3}-\log(2)<0<\frac{7}{10}-\log\left(2\right)$$

$$\frac{2}{3}<\log(2)<\frac{7}{10}$$

A similar set is available with denominators $(1+x)$:

$$\begin{align} \int_0^1 \frac{1}{1+x}dx &= \log(2) \\ \int_0^1 \frac{x}{1+x}dx &= 1-\log(2)\\ \frac{1}{2}\int_0^1 \frac{x^2(1-x)}{1+x} dx &= \log(2)-\frac{2}{3} \\ \frac{1}{2}\int_0^1 \frac{x^5(1-x)}{1+x} dx &= \frac{7}{10}-\log(2) \end{align}$$

and series versions are given by

$$\begin{align} \log(2)-\frac{2}{3} &= \sum_{k=1}^\infty \frac{1}{(2k+1)(2k+2)(2k+3)} \\ \frac{7}{10}-\log(2) &= \sum_{k=2}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)} \\ \end{align} $$