Prove that: $\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$ 
Prove that:
$$\sqrt [3]{36}<\ln 28<\sqrt [3]{37}$$

This inequality is the result of an integral representation/inequality.
I lost access to the article that mentioned this inequality.  Now I want to prove it myself.  Using series is the standard way.  However, this involves heavy calculations, so it's almost useless. We know that,
$$\ln 28=2\ln 2+\ln 7$$
From here, it is necessary to evaluate the $\ln2$ and $\ln 7$ numbers independently. More work is required.

*

*If you've seen this inequality before, please share the solution with us.  Presumably this will be the Integral representation.


*If you have made any solution that requires as little computation as possible, please share it with us.


*The question is open to all solutions involving elementary techniques or advanced techniques.
Less computational solutions are preferred.
 A: For the left inequality:
We have
$$\ln 28 = 5 \ln 2 - \ln(8/7)
\ge 5\ln 2 - 1/7$$
where we have used $\ln(1 + u) \le u$ for all $u \ge 0$.
From
$$0 \le \int_0^1 \frac{x^2(1 - x)^3}{1 + x}\,\mathrm{d} x = 8\ln 2 - \frac{83}{15},$$
we have
$\ln 2 \ge \frac{83}{120}$.
(Note: Alternatively,
we may use $\ln \frac{1+x}{1-x}
\ge 2x + \frac23 x^3$ ($3$-th order Taylor approximation) to get
$\ln 2 \ge \frac{56}{81}$ by letting $x = 1/3$.)
Thus, we have
$$\ln 28 \ge 5 \cdot \frac{83}{120} - 1/7 = \frac{557}{168}.$$
Let $A = \sqrt[3]{36}$ and $B = 10/3$. Then $A < B$. We have
$$A = B - \frac{B^3 - A^3}{B^2 + BA + A^2} \le B - \frac{B^3 - A^3}{3B^2} = \frac{743}{225}.$$
(Note: $\sqrt[3]{36} = 3\sqrt[3]{1 + 1/3} \le 3(1 + 1/9) = 10/3$ by Bernoulli inequality.)
Since $\frac{743}{225} < \frac{557}{168}$, we have $36^{1/3} < \ln 28$.
$\phantom{2}$
For the right inequality:
Using $37 \cdot 27 = 999$, we have
$$\sqrt[3]{37} = \sqrt[3]{999/27} = \frac{10}{3}\sqrt[3]{999/1000}
\ge \frac{10}{3} \cdot \frac{3 \cdot 999/1000}{2\cdot 999/1000 + 1} = \frac{4995}{1499}$$
where we have used
$$u^{1/3} = \frac{u}{\sqrt[3]{u \cdot u \cdot 1}} \ge \frac{u}{(u+u+1)/3} = \frac{3u}{2u + 1}.$$
It suffices to prove that
$$\ln 28 < \frac{4995}{1499}.$$
I don't have a nice approach to deal with it.
Hope it helps.
A: I am trying an approximation with the simple tools...
$$\ln28=3\left(\ln(1+\frac{1}{2})-\ln(1-\frac{1}{2})\right)+\ln(1+\frac{1}{27})\approx3\sum_{n=0}^{N}\frac{1}{(2n+1)2^{2n}}+\sum_{m=1}^{M}\frac{(-1)^{m+1}}{m\;27^{m}}.$$
Now, I assume that $4$ arithmetic operation calculator can be used:
Lower bound is easy: $N=3, M=2\implies \ln28>3.33\implies \ln^328>36.926037>36.$
Upper bound: $N=5, M=3\implies \ln28<3.33213+3\sum_{n=6}^{\infty}\frac{1}{(2n+1)2^{2n}}\implies \ln28<3.33213+\frac{3}{13}\sum_{n=6}^{\infty}\frac{1}{4^{n}}\implies\ln28<3.33221\implies\ln^328<36.9997<37.$
