Integral $\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$ 
Calculate the following integral:
  \begin{equation}
\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx
\end{equation}


I am having trouble to calculate the integral. I tried to use by parts method but it didn't help. Wolfram Alpha gives me $0$ as the answer but I don't know how to get it. I also tried to search the similar question here and I got this:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx,$$
yet it didn't help much. Besides, I don't understand the answers there. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

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
