A definite integral involving the product of a rational function and a logarithm We have to do the following integral.
$$\int_1^{\frac{1+\sqrt{5}}{2}}\frac{x^2+1}{x^4-x^2+1}\ln\left(1+x-\frac{1}{x}\right)dx$$
I tried it a lot.
I substitute $t=1+x-(1/x)$, $dt=1+(1/x^2)$
But then I stuck at
$$\int\limits_{1}^{2} \frac{\ln(t)}{(t-1)^{2} + 1} \mathrm{d}t$$
But now how to proceed.
 A: 
Let $I$ denote the integral
$$I:=\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\mathrm{d}x,$$
with $\phi$ of course being the golden ratio, $\phi=\frac{1+\sqrt{5}}{2}$.
As a numerical approximation, we find
$$I\approx0.272198.$$

Try substituting instead
$$\frac12\left(x-\frac{1}{x}\right)=t,$$
$$\implies\frac12\left(1+\frac{1}{x^{2}}\right)\,\mathrm{d}x=\mathrm{d}t.$$
Taking the square, we have
$$\frac14\left(x^{2}-2+\frac{1}{x^{2}}\right)=t^{2}.$$
Then,
$$\begin{align}
I
&=\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\mathrm{d}x\\
&=\int_{1}^{\phi}\frac{2x^{2}}{x^{4}-x^{2}+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\frac{x^{2}+1}{2x^{2}}\,\mathrm{d}x\\
&=\int_{1}^{\phi}\frac{2}{\left(x^{2}-2+x^{-2}\right)+1}\ln{\left(1+x-\frac{1}{x}\right)}\,\frac12\left(1+\frac{1}{x^{2}}\right)\,\mathrm{d}x\\
&=\int_{0}^{\frac12}\frac{2}{4t^{2}+1}\ln{\left(1+2t\right)}\,\mathrm{d}t;~~~\small{\left[\frac12\left(x-\frac{1}{x}\right)=t\right]}\\
&=\int_{0}^{1}\frac{\ln{\left(1+u\right)}}{u^{2}+1}\,\mathrm{d}u;~~~\small{\left[2t=u\right]}.\\
\end{align}$$

Expressing the logarithmic term as an integral, $I$ becomes a double integral. Changing the order of integration, we obtain
$$\begin{align}
I
&=\int_{0}^{1}\frac{\ln{\left(1+x\right)}}{1+x^{2}}\,\mathrm{d}x\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{1+x^{2}}\int_{0}^{1}\mathrm{d}t\,\frac{x}{1+xt}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{x}{\left(1+x^{2}\right)\left(1+xt\right)}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)\left(1+tx\right)}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\left[\frac{t+x}{\left(1+t^{2}\right)\left(1+x^{2}\right)}-\frac{t}{\left(1+t^{2}\right)\left(1+tx\right)}\right];~~~\small{P.F.D.}\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\int_{0}^{1}\mathrm{d}x\,\left[\frac{t+x}{\left(1+x^{2}\right)}-\frac{t}{\left(1+tx\right)}\right]\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\left[\int_{0}^{1}\mathrm{d}x\,\frac{t}{\left(1+x^{2}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{t}{\left(1+tx\right)}\right]\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\left[\frac{\pi}{4}t+\frac12\ln{\left(2\right)}-\ln{\left(1+t\right)}\right]\\
&=\int_{0}^{1}\frac{\mathrm{d}t}{\left(1+t^{2}\right)}\left[\frac{\pi}{4}t+\frac12\ln{\left(2\right)}\right]-\int_{0}^{1}\frac{\ln{\left(1+t\right)}}{1+t^{2}}\,\mathrm{d}t\\
&=\frac{\pi}{4}\ln{\left(2\right)}-I.\\
\end{align}$$
Thus,
$$I=\frac{\pi}{8}\ln{\left(2\right)}.\blacksquare$$
A: Hint. We have that
$$
\int_1^2\frac{\log(t)}{(t-1)^2+1}dt
=\int_0^1\frac{\log(1+v)}{v^2+1}dv
=\int_0^{\pi/4}\log(1+\tan(u))\,du\\
=\int_0^{\pi/4}\log(\cos(u)+\sin(u))\,du-\int_0^{\pi/4}\log(\cos(u))\,du\\
=\int_0^{\pi/4}\log(\sqrt{2}\cos(\pi/4-u)))\,du-\int_0^{\pi/4}\log(\cos(u))\,du.$$
Now work on the first integral and you will get the answer.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\mc{J} & = \int_{1}^{\pars{1 + \root{5}}/2}{x^{2} + 1 \over x^{4} - x^{2} + 1}\,
\ln\pars{1 + x - {1 \over x}}\,\dd x
\\[5mm] & =
\int_{1}^{\pars{1 + \root{5}}/2}{1 \over x^{2} - 1 + 1/x^{2}}\,
\ln\pars{1 + x - {1 \over x}}\,\pars{1 + {1 \over x^{2}}}\dd x
\\[5mm] & =
\int_{1}^{\pars{1 + \root{5}}/2}{1 \over \pars{x - 1/x}^{2} + 1}\,
\ln\pars{1 + x - {1 \over x}}\,\pars{1 + {1 \over x^{2}}}\dd x
\\[5mm] \stackrel{t\ =\ x - 1/x}{=} &\
\int_{0}^{1}{\ln\pars{1 + t} \over t^{2} + 1}\,\dd t
\,\,\,\stackrel{t\ =\ \tan\pars{\theta}}{=}\,\,\,\
\underbrace{\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta}_{\ds{=\ \mc{J}}}\ =\
\int_{-\pi/4}^{0}\ln\pars{1 + {\tan\pars{\theta} + 1 \over 1 - \tan\pars{\theta}}}\,\dd\theta
\\[5mm] & =
{1 \over 4}\,\pi\ln\pars{2} -
\int_{-\pi/4}^{0}\ln\pars{1 - \tan\pars{\theta}}\,\dd\theta =
\color{#f00}{{1 \over 4}\,\pi\ln\pars{2}} -
\underbrace{\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta}
_{\ds{=\ \mc{J}}}
\end{align}

$$
\mc{J}  = \int_{1}^{\pars{1 + \root{5}}/2}{x^{2} + 1 \over x^{4} - x^{2} + 1}\,
\ln\pars{1 + x - {1 \over x}}\,\dd x =
{\mc{J} + \mc{J} \over 2} = {\pi\ln\pars{2}/4 \over 2} =
\bbx{\ds{{1 \over 8}\,\pi\ln\pars{2}}}
$$
