<s>[INST] An integral and series to prove that $\log(5)>\frac{8}{5}$ A relationship between $\log(5) \approx 1.6094$ and $\dfrac{3}{2}=1.5$ can be justified by the harmonic approximation $$\log(5) \approx H_2=1+\frac{1}{2}=\frac{3}{2}$$
that can be obtained by regrouping Lehmer's logarithm
$$\log(5) = \sum_{k=0}^\infty \left(\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}-\frac{4}{5k+5}\right)$$
symmetrically around the negative term
$$\log(5)=\frac{3}{2}+\sum_{k=1}^\infty \left( \frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2} \right)$$
The corresponding integral is
$$\log(5)-\frac{3}{2}=\int_0^1 \frac{x^2(1-x)(1+3x+x^2)}{1+x+x^2+x^3+x^4}\:dx$$
(answer https://math.stackexchange.com/a/1656356/134791 by Olivier Oloa)
This is a direct proof that $\log(5)>\dfrac{3}{2}$ because the integrand is non-negative in $(0,1)$.
However, $\dfrac{8}{5}=1.6$ would be a closer approximation using small numbers, so the natural question is:

What are Dalzell-type integral and series for $\log(5)-\dfrac{8}{5}$? 

 [/INST] Here is an initial solution. 
Take two truncations of the series
$$\sum_{k=1}^\infty \left(\frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2}\right)$$
that evaluate to $\log(5)-r_1$ and $\log(5)-r_2$ such that
$$r_1<\frac{8}{5}<r_2$$
and build a linear combination of the truncations
$$\alpha r_1+(1-\alpha)r_2=\frac{8}{5}$$
with $0<\alpha<1$.
Because of linearity, the resulting series will have positive terms by construction (when combined properly) and the corresponding integral has positive integrand.

$$\log(5)=\frac{8}{5}+\frac{1}{287} \int_0^1 \frac{x^7(1-x)(1+3x+x^2)(89+198x^{5})}{1+x+x^2+x^3+x^4} dx$$

An acceptable solution should have smaller degree in the numerator.
</s>