<s>[INST] Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from $\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$ 
Show that if $a>1$ then $\log a - \int_a^{a+1} \log x dx$ differs from
  $\frac{-1}{2a}$ by less than $\frac{1}{6a^2}$.

For some $\theta$ between $a-1$ and $a$ and odd $n\in\mathbb{N}$ we have equality:
$$\log a - \int_a^{a+1} \log x dx=\log a - \int_{a-1}^atdt+\int_{a-1}^a \frac{t^2}{2}dt-\int_{a-1}^a \frac{t^3}{3}dt+ \ldots -\int_{a-1}^a \frac{\theta^{n+1}}{n+1}dt$$
I don't know how to move further. Any hint please?
 [/INST] Direct solution
$\int_a^{a+1} \ln(x)\ dx = [ x \ln(x) - x ]_a^{a+1}$
$ = (a+1) \ln(a+1) - a \ln(a) - 1$
$ = \ln(a) + (a+1) \ln(1+\frac{1}{a}) - 1$
$ = \ln(a) + (a+1) ( \frac{1}{a} - \frac{1}{2a^2} + \frac{1}{3a^3} - \frac{1}{4a^4} + \cdots ) - 1$
$ = \ln(a) + \frac{1}{2a} - \frac{1}{6a^2} + \frac{1}{12a^3} - \cdots$ [which is clearly an alternating series since $a \ge 1$].
Thus you get the answer immediately, and much more, since the general term in the asymptotic expansion is trivial to find.
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