Alternatives or generalisations of $\int_0^1 \frac{\ln \left(x^2-x+1\right)}{x \ln x} d x$

Question

The beautiful result of the integral $$ \int_0^1 \frac{\ln \left(x^2-x+1\right)}{x \ln x} d x=\ln 2\ln3 $$ attracts me to tackle it. Noting that $x^3+1=(x+1)(x^2-x+1)$, we can split the integral into two parts as:

$$ \begin{aligned} \int_0^1 \frac{\ln \left(x^2-x+1\right)}{x \ln x} d x & =\int_0^1 \frac{\ln \left(x^3+1\right)-\ln(x+1)}{x \ln x} d x=J(3)-J(1), \end{aligned} $$ where $J(a)=\int_0^1 \frac{\ln \left(x^a+1\right)-\ln \left(x+1\right)}{x \ln x} d x$ whose derivative w.r.t. $a$ is

$$ \begin{aligned} J^{\prime}(a) & =\int_0^1 \frac{x^a \ln x}{x \ln x\left(x^a+1\right)} d x \\ & =\int_0^1 \frac{x^{a-1}}{x^a+1} d x \\ & =\frac{1}{a}\left[\ln \left(x^a+1\right)\right]_0^1 \\ & =\frac{1}{a}\ln 2 \end{aligned} $$ Integrating back yields $$ \int_0^1 \frac{\ln \left(x^2-x+1\right)}{x \ln x} d x =J(3)-J(1)=\int_1^3 J^{\prime}(a) da =\ln 2 \int_1^3 \frac{1}{a} d a =\ln 2 \ln 3 $$

Generalisation 1

For any natural number $n$, $$ \begin{aligned} I_n & =\int_0^1 \frac{\ln \left(x^{2n}-x^{2 n-1}+x^{2 n-2}-\cdots+1\right)}{x \ln x} d x \\ & =\int_0^1 \frac{\ln \left(x^{2 n+1}+1\right)-\ln (x+1)}{x \ln x} d x\\&= \int_1^{2 n+1} \frac{1}{a} \ln 2 d a\\&=\ln 2 \ln (2 n+1) \end{aligned} $$

For example,

$$ \int_0^1 \frac{\ln \left(x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1\right)}{x \ln x}dx=\ln 2\ln 9 $$

Generalisation 2

For any natural number $n$, $$ \int_0^1 \frac{\ln (x^{2n}-x^n+1)}{x \ln x} d x = \int_0^1 \frac{\ln \left(x^{3 n}+1\right)-\ln \left(x^n+1\right)}{x \ln x} d x =\int_n^{3 n} \frac{1}{a} \ln 2 d a =\ln 2 \ln 3 $$ which is independent of the choice of $n$.

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My question

Are there any alternatives or generalisations of the integral?

Your comments and alternatives/generalisations are highly appreciated.

Answer 1

Here is a result I found (which covers all OP's generalisations):

$$\forall n\ge2, V_n:=\int_{0}^{1}\left(\frac{\ln\Phi_n(x)}{x\ln x}+\frac{\Lambda_1(n)}{1-x}\right)dx=\frac{1}{2}\Lambda_2(n)$$ Where $\Phi_n(x)$ is the $n$-th cyclotomic polynomial, $\Lambda_{k}(n)$ is the generalized Von Mangoldt function.

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Proof:
First, consider the integral
$$I(s)=\int_{0}^{1}\left(\frac{\ln(1-x^s)}{x\ln x}+\frac{\ln s}{1-x}-\frac{\ln(1-x)}{\ln x}\right)dx$$ Differentiate w.r.t. $s$ $$I'(s)=\int_{0}^{1}\left(\frac{-x^{s-1}}{1-x^{s}}+\frac{1}{s(1-x)}\right)dx=\int_{0}^{1}\sum_{k=0}^{\infty}\left(\frac{1}{s}x^k-x^{sk+s-1}\right)dx$$ $$=\lim_{x\to 1^{-}}\sum_{k=0}^{\infty}\frac{x^{k+1}-x^{s(k+1)}}{s(k+1)}=\frac{1}{s}\lim_{x\to 1^{-}}\ln\left(\frac{1-x^s}{1-x}\right)=\frac{\ln s}{s}$$ Now, integrate to get $I(s)$ $$I(s)=I(1)+\int_{1}^{s}I'(t)dt=I(1)+\frac{1}{2}\ln^2s$$ Since $n\ge2,\sum_{d|n}\mu(d)=0$ and $\Phi_n(x)=\prod_{d|n}(1-x^d)^{\mu(\frac{n}{d})}$ , we have: $$V_n=\int_{0}^{1}\left(\frac{\ln\Phi_n(x)}{x\ln x}+\frac{\Lambda_1(n)}{1-x}\right)dx=\sum_{d|n}\mu\left(\frac{n}{d}\right)I(d)$$ $$=\sum_{d|n}\mu\left(\frac{n}{d}\right)\left(I(1)+\frac{1}{2}\ln^2 d\right)=\frac{1}{2}\Lambda_2(n)$$ as desired. We can also define $V_1:=I(1)=\int_{0}^{1}\frac{\psi^{(0)}(x+1)+\gamma}{x}dx=1.2577468869...$

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Example: $$V_6=\int_{0}^{1}\left(\frac{\ln\Phi_6(x)}{x\ln x}+\frac{\Lambda_1(6)}{1-x}\right)dx=\int_{0}^{1}\frac{\ln(x^2-x+1)}{x\ln x}dx$$ $$=\frac{1}{2}\Lambda_2(6)=\frac{1}{2}\sum_{d|6}\mu\left(\frac{6}{d}\right)\ln^2 d=\ln(2)\ln(3)$$ For a bonus (see this answer), we can calculate $\Lambda_{2}(n)$ as follow: $$\Lambda_{2}(n)=\sum_{d|n}\Lambda_{1}(d)\Lambda_{1}\left(\frac{n}{d}\right)$$ Finally, there is a nice example that I think it's worth mentioning (proof left as exercise) $$\int_{0}^{1}\frac{\ln(x^8+x^7-x^5-x^4-x^3+x+1)}{x\ln x}dx=0$$

Answer 2

The integral $$ \color{Indigo} I := \int_0^1 \frac{\ln(x^2-x+1)}{x \ln x} dx $$ evaluates to the elegant result $\color{ForestGreen} I = \ln 2 \ln 3$. This was demonstrated by the OP by first noting the algebraic identity $x^2-x+1 = \frac{x^3+1}{x+1}$ and then defining an auxiliary function: $$ {\color{SteelBlue}J}({\color{Teal}k}) := \int_0^1 \frac{\ln(x^{\color{Teal}k}+1) - \ln(x+1)}{x \ln x} dx $$ Differentiating ${\color{SteelBlue}J}({\color{Teal}k})$ with respect to $\color{Teal}k$: $$ {\color{SteelBlue}J}'({\color{Teal}k}) = \frac{d}{d{\color{Teal}k}} \int_0^1 \frac{\ln(x^{\color{Teal}k}+1) - \ln(x+1)}{x \ln x} dx = \int_0^1 \frac{1}{x \ln x} \cdot \frac{x^{\color{Teal}k} \ln x}{x^{\color{Teal}k}+1} dx = \int_0^1 \frac{x^{{\color{Teal}k}-1}}{x^{\color{Teal}k}+1} dx $$ Let $u=x^{\color{Teal}k}$, which implies $du = {\color{Teal}k}x^{{\color{Teal}k}-1}dx$. The limits of integration $(0,1)$ remain unchanged for $u$. $$ {\color{SteelBlue}J}'({\color{Teal}k}) = \int_0^1 \frac{1}{{\color{Teal}k}(u+1)} du = \frac{1}{\color{Teal}k} \big[\ln(u+1)\big]_0^1 = \frac{1}{\color{Teal}k} (\ln(1+1) - \ln(0+1)) = \color{ForestGreen}{\frac{\ln 2}{\color{Teal}k}} $$ Integrating ${\color{SteelBlue}J}'({\color{Teal}k})$ with respect to $\color{Teal}k$ from $1$ to $\color{Teal}k$: $$ {\color{SteelBlue}J}({\color{Teal}k}) - {\color{SteelBlue}J}(1) = \int_1^{\color{Teal}k} \frac{\ln 2}{s} ds = \ln 2 [\ln s]_1^{\color{Teal}k} = \ln 2 (\ln {\color{Teal}k} - \ln 1) = \ln 2 \ln {\color{Teal}k} $$ Since ${\color{SteelBlue}J}(1) = \int_0^1 \frac{\ln(x^1+1) - \ln(x+1)}{x \ln x} dx = \int_0^1 \frac{0}{x \ln x} dx = 0$, we have: $$ {\color{SteelBlue}J}({\color{Teal}k}) = \color{ForestGreen}{\ln 2 \ln {\color{Teal}k}} $$ The original integral $\color{Indigo}I$ corresponds to ${\color{SteelBlue}J}(3)$: $$ \int_0^1 \frac{\ln\left(\frac{x^3+1}{x+1}\right)}{x \ln x} dx = {\color{SteelBlue}J}(3) = \color{ForestGreen}{\ln 2 \ln 3} $$

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OP's Initial Generalizations

The OP elegantly extended this result with two initial generalizations:

1. For the geometric sum $\sum_{j=0}^{2{\color{Teal}n}} (-x)^j = \frac{x^{2{\color{Teal}n}+1}+1}{x+1}$: $$ \color{Indigo}{\int_0^1 \frac{\ln \left(\sum_{j=0}^{2{\color{Teal}n}} (-x)^j\right)}{x \ln x} d x} = \int_0^1 \frac{\ln \left(\frac{x^{2{\color{Teal}n}+1}+1}{x+1}\right)}{x \ln x} d x = {\color{SteelBlue}J}(2{\color{Teal}n}+1) = \bbox[8px, #F0F8FF, border:2px solid #4682B4]{\color{ForestGreen}{\ln 2 \ln (2{\color{Teal}n}+1)}} $$

2. For expressions of the form $x^{2{\color{Teal}n}}-x^{\color{Teal}n}+1 = \frac{x^{3{\color{Teal}n}}+1}{x^{\color{Teal}n}+1}$: $$ \color{Indigo}{\int_0^1 \frac{\ln (x^{2{\color{Teal}n}}-x^{\color{Teal}n}+1)}{x \ln x} d x} = \int_0^1 \frac{\ln \left(\frac{x^{3{\color{Teal}n}}+1}{x^{\color{Teal}n}+1}\right)}{x \ln x} d x $$ This can be expressed as ${\color{SteelBlue}J}(3{\color{Teal}n}) - {\color{SteelBlue}J}({\color{Teal}n})$: $$ {\color{SteelBlue}J}(3{\color{Teal}n}) - {\color{SteelBlue}J}({\color{Teal}n}) = \ln 2 \ln(3{\color{Teal}n}) - \ln 2 \ln {\color{Teal}n} = \ln 2 (\ln(3{\color{Teal}n}) - \ln {\color{Teal}n}) = \ln 2 \ln\left(\frac{3{\color{Teal}n}}{\color{Teal}n}\right) = \bbox[8px, #F0F8FF, border:2px solid #4682B4]{\color{ForestGreen}{\ln 2 \ln 3}} $$

We can build upon this foundation to explore even broader generalizations.

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Core Generalization (Generalization 3)

Let's define a more encompassing function for $\color{Teal}a, \color{Teal}b, \color{Teal}c > 0$: $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) := \int_0^1 \frac{\ln(x^{\color{Teal}a}+{\color{Teal}c}) - \ln(x^{\color{Teal}b}+{\color{Teal}c})}{x \ln x} dx = \color{Indigo}{\int_0^1 \frac{\ln\left(\frac{x^{\color{Teal}a}+{\color{Teal}c}}{x^{\color{Teal}b}+{\color{Teal}c}}\right)}{x \ln x} dx} $$ To evaluate this, consider an auxiliary function ${\color{Navy}F}({\color{Teal}s}; {\color{Teal}\beta}, {\color{Teal}\gamma}) := \int_0^1 \frac{\ln(x^{\color{Teal}s}+{\color{Teal}\gamma}) - \ln(x^{\color{Teal}\beta}+{\color{Teal}\gamma})}{x \ln x} dx$. Our integral ${\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c})$ is ${\color{Navy}F}({\color{Teal}a}; {\color{Teal}b},{\color{Teal}c})$. Note that ${\color{Navy}F}({\color{Teal}b}; {\color{Teal}b},{\color{Teal}c})=0$. Differentiating ${\color{Navy}F}({\color{Teal}s};{\color{Teal}b},{\color{Teal}c})$ with respect to $\color{Teal}s$: $$ \frac{\partial {\color{Navy}F}({\color{Teal}s};{\color{Teal}b},{\color{Teal}c})}{\partial {\color{Teal}s}} = \int_0^1 \frac{1}{x \ln x} \frac{\partial}{\partial {\color{Teal}s}} \left( \ln(x^{\color{Teal}s}+{\color{Teal}c}) \right) dx = \int_0^1 \frac{1}{x \ln x} \frac{x^{\color{Teal}s} \ln x}{x^{\color{Teal}s}+{\color{Teal}c}} dx = \int_0^1 \frac{x^{{\color{Teal}s}-1}}{x^{\color{Teal}s}+{\color{Teal}c}} dx $$ Perform the substitution $u=x^{\color{Teal}s}$, so $du = {\color{Teal}s} x^{{\color{Teal}s}-1} dx$. The limits $(0,1)$ for $x$ correspond to $(0,1)$ for $u$. $$ \frac{\partial {\color{Navy}F}({\color{Teal}s};{\color{Teal}b},{\color{Teal}c})}{\partial {\color{Teal}s}} = \int_0^1 \frac{1}{{\color{Teal}s}(u+{\color{Teal}c})} du = \frac{1}{\color{Teal}s} \big[\ln(u+{\color{Teal}c})\big]_0^1 = \frac{1}{\color{Teal}s} (\ln(1+{\color{Teal}c}) - \ln {\color{Teal}c}) = \color{ForestGreen}{\frac{1}{\color{Teal}s} \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right)} $$ Now, integrate this derivative from $s={\color{Teal}b}$ to $s={\color{Teal}a}$: $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) = {\color{Navy}F}({\color{Teal}a};{\color{Teal}b},{\color{Teal}c}) - {\color{Navy}F}({\color{Teal}b};{\color{Teal}b},{\color{Teal}c}) = \int_{\color{Teal}b}^{\color{Teal}a} \frac{1}{s} \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) ds $$ Since ${\color{Navy}F}({\color{Teal}b};{\color{Teal}b},{\color{Teal}c})=0$: $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) = \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) \int_{\color{Teal}b}^{\color{Teal}a} \frac{1}{s} ds = \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) [\ln s]_{\color{Teal}b}^{\color{Teal}a} $$ This yields the general result: $$ \bbox[15px, #E8F5E9, border:4px groove #388E3C]{\color{MidnightBlue}{ \int_0^1 \frac{\ln\left(\frac{x^{\color{Teal}a}+{\color{Teal}c}}{x^{\color{Teal}b}+{\color{Teal}c}}\right)}{x \ln x} dx = \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) \ln\left(\frac{\color{Teal}a}{\color{Teal}b}\right) }} $$

Connections to Previous Results:

• The OP's original integral is ${\color{SteelBlue}K}(3,1,1) = \ln\left(\frac{1+1}{1}\right)\ln\left(\frac{3}{1}\right) = \color{ForestGreen}{\ln 2 \ln 3}$.
• The OP's Generalization 1 is ${\color{SteelBlue}K}(2{\color{Teal}n}+1, 1, 1) = \ln\left(\frac{1+1}{1}\right)\ln\left(\frac{2{\color{Teal}n}+1}{1}\right) = \color{ForestGreen}{\ln 2 \ln(2{\color{Teal}n}+1)}$.
• The OP's Generalization 2 is ${\color{SteelBlue}K}(3{\color{Teal}n}, {\color{Teal}n}, 1) = \ln\left(\frac{1+1}{1}\right)\ln\left(\frac{3{\color{Teal}n}}{\color{Teal}n}\right) = \color{ForestGreen}{\ln 2 \ln 3}$.

Illustrative Example: $$ \color{Indigo}{\int_0^1 \frac{\ln\left(\frac{x^4+2}{x^2+2}\right)}{x \ln x} dx} = {\color{SteelBlue}K}(4,2,2) = \ln\left(\frac{1+2}{2}\right) \ln\left(\frac{4}{2}\right) = \bbox[10px, #E1F5FE, border:3px groove #0288D1]{\color{ForestGreen}{\ln\left(\frac{3}{2}\right) \ln 2}} $$

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Alternative Derivation via Series Expansion

For $c \ge 1$ (ensuring convergence of the series for $x \in [0,1]$), we can employ the Taylor series for $\color{Navy}{\ln(1+y) = \sum_{k=1}^\infty (-1)^{k-1} \frac{y^k}{k}}$. \begin{align*} \ln(x^{\color{Teal}a}+{\color{Teal}c}) - \ln(x^{\color{Teal}b}+{\color{Teal}c}) &= \ln\left({\color{Teal}c}\left(1+\frac{x^{\color{Teal}a}}{\color{Teal}c}\right)\right) - \ln\left({\color{Teal}c}\left(1+\frac{x^{\color{Teal}b}}{\color{Teal}c}\right)\right) \\ &= \left(\ln {\color{Teal}c} + \ln\left(1+\frac{x^{\color{Teal}a}}{\color{Teal}c}\right)\right) - \left(\ln {\color{Teal}c} + \ln\left(1+\frac{x^{\color{Teal}b}}{\color{Teal}c}\right)\right) \\ &= \ln\left(1+\frac{x^{\color{Teal}a}}{\color{Teal}c}\right) - \ln\left(1+\frac{x^{\color{Teal}b}}{\color{Teal}c}\right) \\ &= \color{MidnightBlue}{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k {\color{Teal}c}^k} (x^{{\color{Teal}a}k} - x^{{\color{Teal}b}k})}\end{align*} Substituting this into the integral for ${\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c})$: $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) = \int_0^1 \frac{1}{x \ln x} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k {\color{Teal}c}^k} (x^{{\color{Teal}a}k} - x^{{\color{Teal}b}k}) dx $$ Assuming uniform convergence, we can interchange the summation and integration: $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k {\color{Teal}c}^k} \int_0^1 \frac{x^{{\color{Teal}a}k-1} - x^{{\color{Teal}b}k-1}}{\ln x} dx $$ We utilize the Frullani-type integral: $\displaystyle \color{Navy}{\int_0^1 \frac{x^P - x^Q}{\ln x} dx = \ln\left(\frac{P+1}{Q+1}\right)}$. Setting $P={\color{Teal}a}k-1$ and $Q={\color{Teal}b}k-1$: $$ \int_0^1 \frac{x^{{\color{Teal}a}k-1} - x^{{\color{Teal}b}k-1}}{\ln x} dx = \ln\left(\frac{({\color{Teal}a}k-1)+1}{({\color{Teal}b}k-1)+1}\right) = \ln\left(\frac{{\color{Teal}a}k}{{\color{Teal}b}k}\right) = \color{ForestGreen}{\ln\left(\frac{\color{Teal}a}{\color{Teal}b}\right)} $$ Therefore, $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k {\color{Teal}c}^k} \ln\left(\frac{\color{Teal}a}{\color{Teal}b}\right) = \ln\left(\frac{\color{Teal}a}{\color{Teal}b}\right) \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \left(\frac{1}{\color{Teal}c}\right)^k $$ The sum is the Taylor series for $\color{Navy}{\ln(1+1/{\color{Teal}c})}$. $$ {\color{SteelBlue}K}({\color{Teal}a},{\color{Teal}b},{\color{Teal}c}) = \ln\left(\frac{\color{Teal}a}{\color{Teal}b}\right) \ln\left(1+\frac{1}{\color{Teal}c}\right) = \color{ForestGreen}{\ln\left(\frac{\color{Teal}a}{\color{Teal}b}\right) \ln\left(\frac{{\color{Teal}c}+1}{\color{Teal}c}\right)} $$ This confirms the result obtained via differentiation for $c \ge 1$. The differentiation method remains more general, holding for all $c>0$.

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Generalization with Products (Generalization 4)

Consider an integral involving products of such terms: $$ {\color{SteelBlue}L} := \color{Indigo}{\int_0^1 \frac{\ln\left(\frac{\prod_{i=1}^{\color{Teal}M} (x^{{\color{Teal}a}_i}+{\color{Teal}c})}{\prod_{i=1}^{\color{Teal}M} (x^{{\color{Teal}b}_i}+{\color{Teal}c})}\right)}{x \ln x} dx} $$ where $\color{Teal}{a_i, b_i, c > 0}$, and the number of factors in the numerator and denominator products, $\color{Teal}M$, is the same. Using the properties of logarithms, the argument of the logarithm becomes:

$$ \ln\left(\frac{\prod_{i=1}^{\color{Teal}M} (x^{{\color{Teal}a}_i}+{\color{Teal}c})}{\prod_{i=1}^{\color{Teal}M} (x^{{\color{Teal}b}_i}+{\color{Teal}c})}\right) = \sum_{i=1}^{\color{Teal}M} \ln(x^{{\color{Teal}a}_i}+{\color{Teal}c}) - \sum_{i=1}^{\color{Teal}M} \ln(x^{{\color{Teal}b}_i}+{\color{Teal}c}) = \sum_{i=1}^{\color{Teal}M} \left( \ln(x^{{\color{Teal}a}_i}+{\color{Teal}c}) - \ln(x^{{\color{Teal}b}_i}+{\color{Teal}c}) \right) $$ The integral ${\color{SteelBlue}L}$ can then be expressed as a sum of ${\color{SteelBlue}K}$-type integrals:

$$ {\color{SteelBlue}L} = \sum_{i=1}^{\color{Teal}M} \int_0^1 \frac{\ln(x^{{\color{Teal}a}_i}+{\color{Teal}c}) - \ln(x^{{\color{Teal}b}_i}+{\color{Teal}c})}{x \ln x} dx = \sum_{i=1}^{\color{Teal}M} {\color{SteelBlue}K}({\color{Teal}a}_i, {\color{Teal}b}_i, {\color{Teal}c}) $$ Substituting the result for ${\color{SteelBlue}K}({\color{Teal}a}_i, {\color{Teal}b}_i, {\color{Teal}c})$:

$$ {\color{SteelBlue}L} = \sum_{i=1}^{\color{Teal}M} \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) \ln\left(\frac{{\color{Teal}a}_i}{{\color{Teal}b}_i}\right) = \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) \sum_{i=1}^{\color{Teal}M} \ln\left(\frac{{\color{Teal}a}_i}{{\color{Teal}b}_i}\right) $$ Using the logarithm property $\sum \ln x_k = \ln (\prod x_k)$:

$$ \bbox[15px, #E8F5E9, border:4px groove #388E3C]{\color{MidnightBlue}{ \int_0^1 \frac{\ln\left(\frac{\prod_{i=1}^{\color{Teal}M} (x^{{\color{Teal}a}_i}+{\color{Teal}c})}{\prod_{i=1}^{\color{Teal}M} (x^{{\color{Teal}b}_i}+{\color{Teal}c})}\right)}{x \ln x} dx = \ln\left(\frac{1+{\color{Teal}c}}{\color{Teal}c}\right) \ln\left(\frac{\prod_{i=1}^{\color{Teal}M} {\color{Teal}a}_i}{\prod_{i=1}^{\color{Teal}M} {\color{Teal}b}_i}\right) }} $$

Illustrative Examples:

1. Consider the integral: $$ \color{Indigo}{ \int_0^1 \frac{\ln\left(\frac{(x^6+1)(x^2+1)}{(x^3+1)(x^1+1)}\right)}{x \ln x} dx } $$ Here, ${\color{Teal}c}=1$, ${\color{Teal}M}=2$. Numerator powers: ${\color{Teal}a}_1=6, {\color{Teal}a}_2=2$. Denominator powers: ${\color{Teal}b}_1=3, {\color{Teal}b}_2=1$. The result is: $$ \ln\left(\frac{1+1}{1}\right) \ln\left(\frac{{\color{Teal}a}_1 {\color{Teal}a}_2}{{\color{Teal}b}_1 {\color{Teal}b}_2}\right) = \ln 2 \ln\left(\frac{6 \cdot 2}{3 \cdot 1}\right) = \ln 2 \ln\left(\frac{12}{3}\right) = \ln 2 \ln 4 = \ln 2 \cdot (2 \ln 2) = \bbox[10px, #E1F5FE, border:3px groove #0288D1]{\color{ForestGreen}{2(\ln 2)^2}} $$

2. Another intriguing example: $$ \color{Indigo}{ \int_0^1 \frac{\ln\left(\frac{(x^{10}+1)(x^3+1)}{(x^5+1)(x^2+1)}\right)}{x \ln x} dx } $$ Here, ${\color{Teal}c}=1$, ${\color{Teal}M}=2$. Numerator powers: ${\color{Teal}a}_1=10, {\color{Teal}a}_2=3$. Denominator powers: ${\color{Teal}b}_1=5, {\color{Teal}b}_2=2$. The result is: $$ \ln 2 \ln\left(\frac{10 \cdot 3}{5 \cdot 2}\right) = \ln 2 \ln\left(\frac{30}{10}\right) = \bbox[10px, #E1F5FE, border:3px groove #0288D1]{\color{ForestGreen}{\ln 2 \ln 3}} $$ Remarkably, this evaluates to the same value as the original integral!

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These generalizations beautifully illustrate the power and flexibility of the OP's original technique. The crucial insight is the ability to decompose the argument of the logarithm into terms of the form $\frac{x^{\color{Teal}a}+{\color{Teal}c}}{x^{\color{Teal}b}+{\color{Teal}c}}$ or products thereof. This structure is amenable to both differentiation under the integral sign and series expansion methods, leading to these elegant closed-form expressions.

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$\blacksquare$