I am trying to evaluate the integral $$\int^2_{\frac{1}2}\ln\left(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})}\right)dx$$ by separating it as $$\int^2_{\frac{1}2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^2_{\frac{1}2}\ln\ln\left(\left(x-\frac{1}{2}\right)^2+4\right)dx$$
I think the substitution $x=1/u$ should be applied for the first item but I don't know how to go on. Please help. Thank you. (The final answer is $-\frac{3}{2}\ln2$)
Separately
\begin{align} \hspace{-10mm}\int^2_{1/2}\ln\ln(x^2-x+\frac{17}{4})dx\overset{x-\frac12\to x} = \int^{3/2}_{0}\ln\ln(x^2+4)\ dx \\ \end{align} \begin{align} \int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)\overset{x\to \frac1x}{dx} = &\ \frac12 \int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)(1+\frac1{x^2})\overset{x-\frac1x\to x}{dx}\\ =&\ \frac12 \int^{3/2}_{-3/2} \ln \frac12 + \ln\ln(x^2+4)\ dx \\ = &-\frac32 \ln2 +\int^{3/2}_{0} \ln\ln(x^2+4)\ dx \\ \end{align}
Together
$$\int^2_{1/2}\ln\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})}dx=-\frac32 \ln2 $$
$$I=\int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^2_{1/2}\ln\ln\left(\left(x-\frac{1}{2}\right)^2+4\right)dx$$
Let $t=x-\frac{1}2$
$$I=\int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)dx-\int^{3/2}_{0}\ln\ln\left(t^2+4\right)dt=I_1-I_2$$
Let $t=x-\frac{1}x,~~dt=dx-d\left(\frac{1}x\right),~~t^2+4=(x+\frac{1}x)^2$
$$\begin{align} I_2&=\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2\left(dx-d\left(\frac{1}x\right)\right)\\ \\ I_2&=\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2 dx-\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2d\left(\frac{1}x\right)~~~~~\text{let}~u=\frac{1}x\\ \\ I_2&=\int^{2}_{1}\ln\ln\left(x+\frac{1}x\right)^2 dx-\int^{1/2}_{1}\ln\ln\left(u+\frac{1}u\right)^2du\\ \\ I_2&=\int^{2}_{1/2}\ln\ln\left(x+\frac{1}x\right)^2 dx \end{align}$$
$$I=I_1-I_2=\int^{2}_{1/2} \ln\left(\frac{1}2\right) dx=-\frac{3}{2}\ln(2)$$
