Update: Finally, a complete solution. Sorry it took so long.
Split the integral up into 3.
\begin{align}
I
&=-\int^{\sqrt{2}}_1\frac{\log{x}}{x}dx+\int^{\sqrt{2}}_1\frac{\log{((x^2-1)^2+1)}}{x}dx-\int^{\sqrt{2}}_1\frac{\log{((x-1)^2+1)}}{x}dx\\
&=-\frac{1}{8}(\log{2})^2+\frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx-\int^{\sqrt{2}-1}_0\frac{\log(1+x^2)}{1+x}dx
\end{align}
The second integral is rather easy to evaluate.
\begin{align}
\frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx
&=\frac{1}{2}\int^1_0\int^1_0\frac{x^2}{(1+x)(1+ax^2)}dx \ da\tag1\\
&=\frac{1}{2}\int^1_0\frac{1}{1+a}\int^1_0\frac{1}{1+x}+\frac{x-1}{1+ax^2}dx \ da\\
&=\frac{1}{2}\int^1_0\frac{\log{2}}{1+a}+\frac{\log(1+a)}{2a(1+a)}-\underbrace{\frac{\arctan(\sqrt{a})}{\sqrt{a}(1+a)}}_{\text{Let} \ y=\arctan{\sqrt{a}}}da\\
&=\frac{1}{2}\left[(\log{2})^2+\frac{1}{2}\underbrace{\int^1_0\frac{\log(1+a)}{a}da}_{-\operatorname{Li}_2(-1)=\frac{\pi^2}{12}}-\frac{1}{2}\underbrace{\int^1_0\frac{\log(1+a)}{1+a}da}_{\frac{1}{2}(\log{2})^2}-\frac{\pi^2}{16}\right]\\
&=\frac{3}{8}(\log{2})^2-\frac{\pi^2}{96}
\end{align}
The third integral can be evaluated using dilogarithms.
\begin{align}
\int^{\sqrt{2}-1}_0\frac{\log(1+x^2)}{1+x}dx
&=\sum_{r=\pm i}\int^{\sqrt{2}-1}_0\frac{\log(r+x)}{1+x}dx\tag2\\
&=-\sum_{r=\pm i}\int^{\frac{\lambda}{\sqrt{2}}}_{\lambda}\log\left(r-1+\frac{\lambda}{y}\right)\frac{dy}{y}\tag3\\
&=-\sum_{r=\pm i}\int^{\frac{r-1}{\sqrt{2}}}_{r-1}\frac{\log(1+y)}{y}-\frac{1}{y}\log\left(\frac{y}{r-1}\right)dy\tag4\\
&=\frac{1}{4}(\log{2})^2+\sum_{r=\pm i}\mathrm{Li}_2\left(\frac{1-r}{\sqrt{2}}\right)-\mathrm{Li}_2(1-r)\tag5\\
&=\frac{1}{4}(\log{2})^2+\mathrm{Li}_2(e^{i\pi/4})+\mathrm{Li}_2(e^{-i\pi/4})-\mathrm{Li}_2(\sqrt{2}e^{i\pi/4})-\mathrm{Li}_2(\sqrt{2}e^{-i\pi/4})\\
&=\frac{1}{4}(\log{2})^2-\frac{\pi^2}{96}\tag6\\
\end{align}
It follows that
$$I=-\frac{1}{8}(\log{2})^2+\frac{3}{8}(\log{2})^2-\frac{\pi^2}{96}-\frac{1}{4}(\log{2})^2+\frac{\pi^2}{96}=0$$
Explanation
$(1)$: Differentiate under the integral sign
$(2)$: Factorise $1+x^2$, let $r=\pm i$
$(3)$: Let $\displaystyle y=\frac{\lambda}{1+x}$
$(4)$: Let $\lambda=r-1$
$(5)$: Recognise that $\displaystyle\int\frac{\ln(1+y)}{y}dy=-\mathrm{Li}_2(-y)+C$ and $\displaystyle\int\frac{\ln(ay)}{y}dy=\frac{1}{2}\ln^2(ay)+C$
$(6)$: Use the identities here
You can use the identity given by the Euler Beta function
$$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
to state:
$$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$
and by switching the series and the integral:
$$ S = \int_{0}^{1}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx,$$
$$ S = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{1/4-x^2})}{1/4-x^2}dx = 4\int_{0}^{1}\frac{\log(1+\frac{1}{2}\sqrt{1-x^2})}{1-x^2}dx,$$
$$ S = 4\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}d\theta.$$
Now Mathematica gives me $\frac{5\pi^2}{18}$ as an explicit value for the last integral, but probably we are on the wrong path, and we only need to exploit the identity
$$\sum_{k=1}^{+\infty}\frac{1}{k^2\binom{2k}{k}}=\frac{\pi^2}{18}$$
that follows from the Euler acceleration technique applied to the $\zeta(2)$-series. The other "piece" (the $U$-piece in the Marty Cohen's answer) is simply given by the Taylor series of $\arcsin(z)^2$. More details to come.
As a matter of fact, both approaches lead to an answer.
The (Taylor) series approach, as Bhenni Benghorbal shows below, leads to the identity:
$$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right),\tag{1}$$
while the integral approach, as Achille Hui pointed out in the comments, leads to:
$$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}\,d\theta&=&\int_{0}^{1}\log\left(1+\frac{t}{1+t^2}\right)\frac{dt}{t}\\&=&\int_{0}^{1}\frac{\log(1-t^3)-\log(1-t)-\log(1+t^2)}{t}\,dt\\&=&\int_{0}^{1}\frac{-\frac{2}{3}\log(1-t)-\frac{1}{2}\log(1+t)}{t}\,dt\\&=&\frac{1}{6}\sum_{k=1}^{+\infty}\frac{4+3(-1)^k}{k^2}=\frac{1}{6}\left(4-\frac{3}{2}\right)\zeta(2)=\frac{5\pi^2}{72}.\end{eqnarray*}\tag{2}$$
Thanks to both since now this answer may become a reference both for integral-log-ish problems like $(2)$ and for $\Gamma^2$-series like $(1)$.
Update 14-06-2016. I just discovered that this problem can also be solved by computing
$$ \int_{-1}^{1} x^n\, P_n(x)\,dx, $$
where $P_n$ is a Legendre polynomial, through Bonnet's recursion formula or Rodrigues' formula. Really interesting.
Best Answer
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ With $\ds{r \equiv {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$ \begin{align}&\color{#c00000}{ \int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over x^{2} - x + 1}\,\dd x} =\int_{0}^{1}{\ln\pars{\ln\pars{1/x}} \over \pars{x - r}\pars{x - r^{*}}}\,\dd x \\[3mm]&=\int_{0}^{1}\ln\pars{\ln\pars{1/x}} \pars{{1 \over x - r} - {1 \over x - r^{*}}}\,{1 \over r - r^{*}}\,\dd x \\[3mm] & = {1 \over \Im\pars{r}}\,\Im\int_{0}^{1} {\ln\pars{\ln\pars{1/x}} \over x - r}\,\dd x \end{align}
$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$. Derivatives of the PolyLogarithm, respect of the order, can be evaluated from its integral representation.
Also, see Hurwitz Zeta Function.