I'll not bother showing all the integrals converge. They all concern bounded functions continuous in a compact domain $[0,1]$.

Solve:
\begin{equation}
\int_{0}^1x\left \{\frac{1}{x}\right \}\mathrm{d}x
\end{equation}

We begin by remembering that $\left \{\frac{1}{x}\right \} =\frac{1}{x}- \left \lfloor \frac{1}{x} \right \rfloor$:
\begin{equation}
\int_{0}^1x\left \{\frac{1}{x}\right \}\mathrm{d}x = \int_{0}^1\mathrm{d}x-\int_{0}^1 x\left \lfloor \frac{1}{x}\right \rfloor \mathrm{d}x=1-\int_{0}^1 x\left \lfloor \frac{1}{x}\right \rfloor \mathrm{d}x
\end{equation}

Thus, we solve the problem by computing $\int_{0}^1 x\left \lfloor \frac{1}{x}\right \rfloor \mathrm{d}x$. Consider the following sequence:

\begin{equation}
\mathcal{I}_N = \int_{\frac{1}{N}}^1 x\left \lfloor \frac{1}{x}\right \rfloor \mathrm{d}x=\sum_{n=1}^{N-1}\int_{\frac{1}{n+1}}^{\frac{1}{n}}x\left \lfloor \frac{1}{x}\right \rfloor \mathrm{d}x
\end{equation}

However, if $\frac{1}{n+1}<x\leq \frac{1}{n}\Rightarrow n\leq \frac{1}{x}<n+1\Rightarrow \left \lfloor \frac{1}{x}\right \rfloor = n$. This means we can solve each of the parcels by substituting this value:
\begin{align}
\mathcal{I}_N &=\sum_{n=1}^{N-1}\int_{\frac{1}{n+1}}^{\frac{1}{n}}x\left \lfloor \frac{1}{x}\right \rfloor \mathrm{d}x=\mathcal{I}_N =\sum_{n=1}^{N-1}\int_{\frac{1}{n+1}}^{\frac{1}{n}}nx \mathrm{d}x=\sum_{n=1}^{N-1} n \left .\frac{x^2}{2}\right |_{\frac{1}{n+1}}^{\frac{1}{n}}\\
&=\sum_{n=1}^{N-1}\frac{n}{2}\left (\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)=\frac{1}{2}\sum_{n=1}^{N-1}\left ( \frac{1}{n}-\frac{n}{(n+1)^2}\right )\\
&=\frac{1}{2}\sum_{n=1}^{N-1} \left (\frac{1}{n} - \frac{1}{n+1} +\frac{1}{(n+1)^2} \right )=\frac{1}{2}\sum_{n=1}^{N-1} \left (\frac{1}{n} - \frac{1}{n+1}\right ) + \frac{1}{2}\sum_{n=1}^{N-1} \left ( \frac{1}{(n+1)^2} \right )
\end{align}

The first sum is what we call a telescopic sum, and thus can be easily simplified to give us a convenient form for $\mathcal{I}_N$:

\begin{equation}
\mathcal{I}_N = \frac{1}{2}\left ( 1 - \frac{1}{N} \right ) + \frac{1}{2}\sum_{n=2}^{N} \frac{1}{n^2} = \frac{1}{2}\sum_{n=1}^{N} \frac{1}{n^2}-\frac{1}{2N}
\end{equation}

To solve this, we utilize that:

\begin{equation}
\int_{0}^1x\left \{\frac{1}{x}\right \}\mathrm{d}x = 1-\lim_{N\rightarrow \infty} \mathcal{I}_N = 1-\frac{1}{2}\sum_{n=1}^\infty \frac{1}{n^2} = 1-\frac{\pi^2}{12}
\end{equation}