Show that $\log(r)$ minus an integral involving the $\log$ function is less than $\frac{1}{6r^2}$ I have to prove that:

If $r > 1$ then $\log r - \int_{r}^{r+1}\log(x)\, dx$ differs from $-\frac{1}{2r}$ by less than $\frac{1}{6r^2}$.

I am thinking about working out the integral using Taylor Series for log(1+x) at the point 0. This seems like it will work, but when I attempt it I do not get the correct statement. Any hints or alternate proofs that would work?
 A: $$\int_{n}^{n+1}\log(x)\,dx -\log(n) = \int_{n}^{n+1}\log\left(\frac{x}{n}\right)\,dx = \int_{0}^{1}\log\left(1+\frac{x}{n}\right)\,dx \tag{1}$$
but for every $z>-1$:
$$ z-\frac{z^2}{2}\leq\log(1+z)\leq z \tag{2} $$
hence by termwise integration:
$$ \frac{1}{2n}-\frac{1}{6n^2}\leq \int_{n}^{n+1}\log(x)\,dx -\log(n) \leq \frac{1}{2n}.\tag{3}$$
We may also notice that, by integration by parts:
$$\begin{eqnarray*}\int_{0}^{1}\left[\frac{x}{n}-\log\left(1+\frac{x}{n}\right)\right]\,dx&=&\frac{1}{n}-\log\left(1+\frac{1}{n}\right)-\int_{0}^{1}\frac{x^2}{n^2+nx}\,dx\\&=&\color{red}{\int_{0}^{1}\frac{x(1-x)}{n(n+x)}\,dx}\\&\leq&\frac{1}{n^2}\int_{0}^{1}x(1-x)\,dx = \frac{1}{6n^2}\tag{4}  \end{eqnarray*}$$
and that upper bound can be improved up to $\color{red}{\frac{4n+1}{12n^2(2n+1)}}$ by splitting the red integral over $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right]$. Moreover, $\color{red}{\frac{1}{6n^2+6n}}$ can be took as a lower bound.
A: Consider
$$\int_n^{n+1} dx \, f(x) = F(n+1)-F(n)$$
where $F'(x)=f(x)$.  For large $n$, we may write (so long as $F$ is sufficiently well-behaved)
$$\begin{align}F(n+1)-F(n) &= F'(n) (1) + \frac1{2!} F"(n) (1)^2 + \frac1{3!} F"'(n) (1^3) +\cdots \\ &= f(n) + \frac12 f'(n) + \frac16 f"(n)+\cdots\end{align}$$
In your case, $f(x) = \log{x}$ so that
$$\log{n} - \int_n^{n+1} dx \, \log{x} = \frac1{2 n}- \frac1{6 n^2}+ \frac1{12 n^3}+\cdots$$
and-\
$$\left | \left ( \log{n} - \int_n^{n+1} dx \, \log{x}-\frac1{2 n} \right ) \right| = \frac1{6 n^2}- \frac1{12 n^3}+\cdots \le \frac1{6 n^2}$$
