A curious integral involving the fractional part function

$\begin{aligned} I & = \int_0^1 \left \{ \frac 1x \right \}^2 dx \\& = \sum_{k=1}^\infty \int_{\frac 1{k+1}}^\frac 1k \left(\frac 1x - k \right)^2 dx \\ & = \sum_{k=1}^\infty \int_{\frac 1{k+1}}^\frac 1k \left(\frac 1{x^2} - \frac {2k}x + k^2 \right) dx \\ & = \sum_{k=1}^\infty \left[-\frac 1{x} - 2k \ln x + k^2 x \right]_{\frac 1{k+1}}^\frac 1k \\ & = \sum_{k=1}^\infty \left[-k+k+1 - 2k \left(\ln \frac 1k - \ln \frac 1{k+1}\right) + k^2 \left(\frac 1k - \frac 1{k+1} \right) \right] \\ & = \sum_{k=1}^\infty \left[\frac {2k+1}{k+1} - 2k \ln \frac {k+1}k \right] \\ & \approx 0.259005 \end{aligned}$

To finish what Chew-Seong has done, the closed form of the answer should be

$\ln(2\pi)-\gamma-1$

Where $\gamma$ is the Euler-Mascheroni constant.