About the integral $\int_{\frac{1}{2}}^{2}\ln\left(\frac{\ln\left(x+\frac{1}{x}\right)}{\ln\left(x^{2}-x+\frac{17}{4}\right)}\right)dx$

Question

$$I=\int_{\frac{1}{2}}^{2}\ln\left(\frac{\ln\left(x+\frac{1}{x}\right)}{\ln\left(x^{2}-x+\frac{17}{4}\right)}\right)dx$$ Using CAS the answer comes out to be: $$I=-\frac{3}{2}\ln(2)$$

For the Indefinite Integral, Wolfram says it has no solution in terms of standard mathematical functions.

The limits of the Definite Integral do give out the hint of replacing $x \to \frac{1}{x}$, but after trying it, I do not end up with a solvable Integral either. $$I=\int_{\frac{1}{2}}^{2}\ln\left(\frac{\ln\left(x+\frac{1}{x}\right)}{\ln\left(\frac{1}{x^{2}}-\frac{1}{x}+\frac{17}{4}\right)}\right)\frac{1}{x^{2}}dx$$

Adding these two integrals which usually helps does not resolve the problem here either.

Answer 1

$$\int_\frac12^2\ln\left(\frac{\ln\left(x+\frac{1}{x}\right)}{\ln\left(\left(x-\frac12\right)^2+4\right)}\right)dx=\int_\frac12^2\ln\left(\frac{\color{magenta}{2}\ln\left(x+\frac{1}{x}\right)}{\color{magenta}2\ln\left(\left(x-\frac12\right)^2+4\right)}\right)dx$$

$$=-\ln 2 \underbrace{\int_\frac12^2 dx}_{=\frac32}+\underbrace{\int_\frac12^2\ln\left(\frac{\color{red}{\ln\left(\left(x+\frac{1}{x}\right)^2\right)}}{\color{blue}{\ln\left(\left(x-\frac12\right)^2+4\right)}}\right)dx}_{=0}=-\frac32\ln 2$$

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In order to show that the second integral vanishes, we can split the logarithm in two parts.

$$\int_\frac12^2\ln\left(\color{red}{\ln\left(\left(x+\frac1x\right)^2\right)}\right)dx=\int_\frac12^1+\int_1^2\overset{x+\frac1x\to x}=\int_2^\frac52\frac{x\ln(\ln(x^2))}{\sqrt{x^2-4}}dx$$

$$\overset{x^2-4\to x^2}=\int_0^\frac32\ln(\ln(x^2+4))dx\overset{x\to x-\frac12}=\int_\frac12^2\ln\left(\color{blue}{\ln\left(\left(x-\frac12\right)^2+4\right)}\right)dx$$

Answer 2

Note$$I=\int_{\frac{1}{2}}^{2}\ln\frac{\ln\left(x+\frac{1}{x}\right)}{\ln\left(x^{2}-x+\frac{17}{4}\right)}dx=I_2-I_1$$ where

\begin{align} I_1=&\int_{\frac{1}{2}}^{2}\ln \ln\left(x^2-x+\frac{17}4\right)\overset{y=x-\frac12}{dx} =\int_{0}^{\frac32}\ln \ln\left( y^2+4\right)dy\\ \\ I_2=&\int_{\frac{1}{2}}^{2}\ln \ln\left(x+\frac{1}{x}\right)\overset{x\to\frac1x}{dx}\\ =& \ \frac12\int_{\frac{1}{2}}^{2}\ln \ln\left(x+\frac{1}{x}\right)\left(1+\frac1{x^2}\right) \overset{y=x-\frac1x}{dx}\\ =& \ \frac12\int_{-\frac{3}{2}}^{\frac32}\ln \left(\frac12\ln\left(y^2+4\right)\right)dy = \int_{0}^{\frac32}\ln \left(\frac12\ln\left(y^2+4\right)\right)dy\\ \end{align} Thus $$I=\int_0^{\frac32}\ln\frac12=-\frac32\ln2$$