This integral is not equal to zero.
We may obtain the following closed form.
$$
\begin{align}
\int_0^1 \left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\frac{\log(x)}{x}
\mathrm{d}x & = \dfrac{\ln^2(2\pi)}{4}-\dfrac{\gamma^2}{4}+\dfrac{\pi^2}{48}-\dfrac{\gamma_1}{2}-1\tag1 \\\\
\end{align}
$$
where $\left\{x\right\}$ denotes the fractional part of $x$, $\gamma$ denotes the Euler–Mascheroni constant and where $\gamma_{1}$ denotes the Stieltjes constant defined by
$$
\gamma_{1} = \lim_{N \rightarrow \infty}\left(\sum_{k=1}^{N}\frac{\ln k}{k}-\frac{\ln^{2}N}{2} \right).
$$
Consequently, we have the numerical evaluation:
$$
\begin{align}
\int_0^1 \left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\frac{\log(x)}{x}
\mathrm{d}x = \color{red}{0.00}31782279542924256050500... . \tag2
\end{align}
$$
Here is an approach.
Step 1. Let $s$ be a complex number such that $0<\Re{s}<1$. Then
$$
\int_{0}^{1} x^{s-1}\left\{\frac{1}{x}\right\}\mathrm{d}x = -\frac{1}{1-s} -\frac{\zeta(s)}{s}\tag3
$$ where $\left\{x\right\}$ denotes the fractional part of $x$ and where $\zeta$ denotes the Riemann zeta function.
Proof. Let us assume that $0<\Re{s}<1$. We may write
$$
\begin{align}
\int_{0}^{1} x^{s-1}\left\{\frac{1}{x}\right\}\mathrm{d}x & = \sum_{k=1}^{\infty}
\int_{1/(k+1)}^{1/k} x^{s-1}\left\{\frac{1}{x}\right\}\mathrm{d}x \\
& = \sum_{k=1}^{\infty} \int_{k}^{k+1} \left\{x\right\} \frac{\mathrm{d}x}{x^{s+1}} \\
& = \sum_{k=1}^{\infty} \int_{k}^{k+1} (x-k) \frac{\mathrm{d}x}{x^{s+1}} \\
& = \sum_{k=1}^{\infty} \int_{0}^{1}\frac{v}{(v+k)^{s+1}}\mathrm{d}v \\
& = \sum_{k=1}^{\infty} \int_{0}^{1}\left(\frac{1}{(v+k)^{s}}-\frac{k}{(v+k)^{s+1}}\right)\mathrm{d}v \\
& = \sum_{k=1}^{\infty} \left.\left(\frac{1}{(-s+1)(v+k)^{s-1}} +\frac{k}{s(v+k)^s}\right) \right|_{0}^{1} \\
& = -\frac{1}{1-s}-\frac{\zeta(s)}{s}.
\end{align}
$$
Step 2. We have
$$
\int_{0}^{1} x^{s-1}\left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\log(x)\mathrm{d}x = -\frac{1}{(1-s)^2} +\frac{1}{2s^2} +\frac{\zeta(s)}{s^2} -\frac{\zeta'(s)}{s}. \tag4
$$
Using $(3)$, we readily get
$$
\int_{0}^{1} x^{s-1}\left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\mathrm{d}x = -\frac{1}{1-s}-\frac{1}{2s} -\frac{\zeta(s)}{s}
$$
which we differentiate with respect to $s$ to obtain $(4)$.
Step 3. For $s$ near $0$, we take into account the Taylor series expansion of the Riemann $\zeta$ function:
$$
\begin{align}
& \zeta(s) =-\frac12-\dfrac{\ln(2\pi)}{2} s +\left(\dfrac{\gamma^2}{4}-\dfrac{\pi^2}{48}+\ln(2\pi)-\dfrac{\ln^2(2\pi)}{4}+\dfrac{\gamma_1}{2}\right)s^2+\mathcal{O}(s^3)
\\& \zeta'(s) =-\dfrac{\ln(2\pi)}{2} +\left(\dfrac{\gamma^2}{2}-\dfrac{\pi^2}{24}+2\ln(2\pi)-\dfrac{\ln^2(2\pi)}{2}+\gamma_1\right)s+\mathcal{O}(s^2)
\end{align}
$$
and upon letting $s$ tend to $0^+$ in $(4)$ we obtain $(1)$.
Remark: A related result to $(3)$.
Starting with the following identity that can be found on page $95$ Eq $(5)$ of this paper
$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$
Multiply both sides by $\large \frac{\ln(1-x^2)}{x}$ then integrate from $x=0$ to $x=1$ we get
$$\underbrace{\sum_{n=1}^\infty \frac{\overline{H}_n}{n}\int_0^1x^{n-1}\ln(1-x^2)\ dx}_{\large S}$$
$$\small{=I-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x^2)}{x}\ dx}_{\large J}-\ln2\underbrace{\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}\ dx}_{\large K}-\operatorname{Li}_2\left(\frac12\right)\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}\ dx}_{\large -\frac12\zeta(2)}}$$
or
$$I=S+J+\ln2\ K-\frac12\zeta(2)\operatorname{Li}_2\left(\frac12\right)\tag1$$
Evaluating $S$
Notice that
$$\int_0^1 x^{n-1}\ln(1-x^2)\ dx\overset{x^2\to x}{=}\frac12\int_0^1 x^{n/2-1}\ln(1-x)\ dx=-\frac{H_{n/2}}{n}$$
$$\Longrightarrow S=\boxed{-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}}$$
Evaluating $J$
Writing $\ln(1-x^2)=\ln(1-x)+\ln(1+x)$ gives
$$J=\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x)}{x}dx+\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1+x)}{x}dx$$
$$=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln(1-x)\ dx-\frac12\operatorname{Li}_2^2(-x)|_0^1$$
$$=-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac{5}{16}\zeta(4)$$
$$=\boxed{-2\operatorname{Li_4}\left(\frac12\right)+\frac{39}{16}\zeta(4)-\frac74\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$
where we used $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$
Evaluating $K$
Similarly
$$K=\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{\large 2\zeta(3)}+\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{\large -\frac{5}{8}\zeta(3)}=\boxed{\frac{11}8\zeta(3)}$$
where the second integral is evaluated here.
Plug the boxed results in $(1)$ we obtain that
$$I=-2\operatorname{Li}_4\left(\frac12\right)+\frac{29}{16}\zeta(4)-\frac38\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac1{12}\ln^42-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$$
In here we proved
$$\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)$$
$$\Longrightarrow I=-\frac1{8}\ln^42+\ln^22\zeta(2)-3\ln2\zeta(3)+\frac{47}{16}\zeta(4)-3\operatorname{Li}_4\left(\frac12\right)$$
Best Answer
Just to give a different approach, first substitute $u={1+x\over1-x}$ and then integrate by parts:
$$u={1+x\over1-x}\implies x={u-1\over u+1}\implies dx={2du\over(u+1)^2}$$
so
$$\begin{align} \int_0^1\ln\left(1+x\over1-x\right)\,dx &=2\int_1^\infty{\ln u\over(u+1)^2}\,du\\ &={-2\ln u\over(u+1)}\Big|_1^\infty+2\int_1^\infty{1\over u(u+1)}\,du\\ &=(0-0)+2\int_1^\infty\left({1\over u}-{1\over u+1}\right)\,du\\ &=2\ln\left(u\over u+1\right)\Big|_1^\infty\\ &=2(\ln1-\ln(1/2))\\ &=2\ln2 \end{align}$$