Finding $\int_0^1 \left\{\frac1x\right\}\left\{\frac1{1-x}\right\}\left\{1-\frac1x\right\}dx$ where $\{x\}$ is fractional part of $x$ I was able to evaluate this from $0$ to $\frac{1}{2}$ but the middle part becomes too hard to evaluate under $\frac{1}{2}$ to $1$. I only have a high school level of understanding of maths, please make it easier to understand.

$$\int_0^1 \left\{\frac1x\right\}\left\{\frac1{1-x}\right\}\left\{1-\frac1x\right\}\,\mathrm{d}x=?$$

where $\{x\}$ is fractional part of $x$.
I haven't slogged through all the calculations, but I can explain how to do it.  Actually, if you were able to compute the integral from $0$ to $\frac12$, you've got the basic idea.  you just have to modify it to compute the integral from $\frac12$ to $1$.

The basic idea of course is to split $[0,1]$ into intervals on which the integer parts of the various expressions are constant.  

$$\begin{align}
&amp;\left\lfloor\frac1x\right\rfloor=n \iff\\ 
&amp;n\leq \frac1x&lt;n+1\iff\\
&amp;\frac1{n+1}&lt;x\leq\frac1n
\end{align}$$
In particular, the integer part is $1$ for $\frac12&lt;x\leq1$, and of course the integer part of $1-\frac1x$ is $-1$.

Similarly, $$\begin{align}
&amp;\left\lfloor\frac1{1-x}\right\rfloor=n\iff\\
&amp;n\leq\frac1{1-x}&lt;n+1\iff\\
&amp;\frac1{n+1}&lt;1-x\leq\frac1n\iff\\
&amp;-\frac1n\leq x-1&lt;-\frac1{n+1}\iff\\
&amp;\frac{n-1}n\leq x&lt;\frac n{n+1}
\end{align}$$
In particular, the integer part is $1$ for $0\leq x&lt;\frac12$.

Therefore, we need to split the interval $[0,\frac12]$ up into intervals of the form $\left(\frac1{n+1},\frac12\right]$, which I assume you did, and we need to split the interval $\left(\frac12,1\right]$ up into intervals of the form $\left[\frac {n-1}n, \frac n{n+1}\right)$.
Consider the integral

$$I_k = \int_{\frac{1}{k+1}}^\frac{1}{k}\{\frac{1}{x}\}\{1-\frac{1}{x}\}\{{\frac{1}{1-x}}\}dx$$

For this range of $x$, we have

$$\{\frac{1}{x}\} = \frac{1}{x} - k \\
\{1 - \frac{1}{x}\} = k+1 - \frac{1}{x} \\
\{\frac{1}{1-x}\} = \frac{1}{1-x} -1$$

Now, substituting, we have

$$I_k = \int_{\frac{1}{k+1}}^\frac{1}{k}\left(\frac{1}{x} - k\right)\left(k+1 - \frac{1}{x}\right)\left(\frac{1}{1-x} - 1\right)dx$$

You can use partial fractions to decompose this into $\ln$ terms to get $I_k$, and then sum $I_k$ to get $I$