${\displaystyle \int_0^1 \dfrac{1}{x} \cdot \dfrac{1}{1-x} \cdot \left(1-\dfrac{1}{x} \right) \mathrm dx}$ 
Evaluate
$$\displaystyle \int_0^1 \left\{\dfrac{1}{x}\right\} \cdot \left\{\dfrac{1}{1-x}\right\} \cdot \left\{1-\dfrac{1}{x} \right\} \mathrm dx$$
where $\{x\}$ denotes the fractional part of $x$.


Since $\{1 - \frac{1}{x}\} = \{-\frac{1}{x}\} = 1 - \{\frac{1}{x}\}$, we have
\begin{align}
I &= \int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} (1 - \{\tfrac{1}{x}\})\mathrm{d}x \\
&= \int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} \mathrm{d}x
- \int_0^1 \{\tfrac{1}{x}\}^2\{\tfrac{1}{1-x}\} \mathrm{d}x.
\end{align}
From 2.10 and 2.12 (page 101) in [1], we have
$$\int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} \mathrm{d}x = 2\gamma - 1$$
and
$$\int_0^1 \{\tfrac{1}{x}\}^2\{\tfrac{1}{1-x}\} \mathrm{d}x = \frac{5}{2} - \gamma - \ln (2\pi)$$
where $\gamma $ is Euler-Mascheroni constant.
Thus, we have $I = 3\gamma - \frac{7}{2} + \ln (2\pi) \approx 0.069524062$.
Also, I approximated the integral numerically in Maple which resembles the analytic result above.
Reference: 
[1] Ovidiu Furdui, "Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis".
