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