Is it a kind of double integral??

The substitution $x = e^{-y}$ gives $I \; =\; \int_0^1 \frac{x-1}{(x+1)\ln x}\,dx \; =\; \int_0^\infty \frac{(1-e^{-y})e^{-y}}{(1 + e^{-y})y}\,dy \; = \; F(1)$ where $F(u) \; = \; \int_0^\infty \frac{(1 - e^{-y})e^{-uy}}{(1 + e^{-y})y}\,dy \hspace{2cm} u > 0$ The Monotone Convergence Theorem shows that $\lim_{u\to\infty}F(u) =0$ , and we can differentiate under the integral sign to obtain $\begin{aligned} F'(u) & = \; -\int_0^\infty \frac{(1 - e^{-y})e^{-uy}}{1 + e^{-y}}\,dy \; = \; -\int_0^\infty (1 - e^{-y})e^{-uy}\left(\sum_{n=0}^\infty (-1)^n e^{-ny}\right)\,dy \\ & = \; -\sum_{n=0}^\infty (-1)^n \int_0^\infty (1 - e^{-y})e^{-(u+n)y}\,dy \; = \; -\sum_{n=0}^\infty \frac{(-1)^n}{(u+n)(u+n+1)} \end{aligned}$ where the interchange of integration and summation is justified by the Dominated Convergence Theorem. Thus, applying the Dominated Convergence Theorem again, $\begin{aligned} I - F(u) \; =\; F(1) - F(u) & = \; \int_1^u \sum_{n=0}^\infty \frac{(-1)^n}{(v+n)(v+n+1)}\,dv \; =\; \sum_{n=0}^\infty (-1)^n\int_1^u \frac{dv}{(v+n)(v+n+1)} \; =\; \sum_{n=0}^\infty (-1)^n \Big[\ln\left(\frac{v+n}{v+n+1}\right)\Big]_1^u \\ & = \; \sum_{n=0}^\infty (-1)^n \ln\left(\frac{u+n}{u+n+1}\,\frac{n+2}{n+1}\right) \; = \; \sum_{n=0}^\infty \ln\left(\frac{2n+u}{2n+u+1}\,\frac{2(n+1)}{2n+1}\,\frac{2n+u+2}{2n+u+1}\,\frac{2(n+1)}{2n+3}\right) \\ & = \; \sum_{n=0}^\infty \ln\left(\frac{4(n+1)^2}{4(n+1)^2 - 1}\,\frac{(2n+u+1)^2-1}{(2n+u+1)^2}\right) \; =\; \sum_{n=0}^\infty\ln\left(\frac{1 - \frac{1}{(2n+u+1)^2}}{1 - \frac{1}{4(n+1)^2}}\right) \\ & = \; \ln\left(\prod_{n=1}^\infty \left(1 - \frac{1}{4n^2}\right)^{-1}\right) + \ln\left(\prod_{n=0}^\infty \left(1 - \frac{1}{(2n+u+1)^2}\right)\right) \\ & = \; \ln\left(\tfrac{\pi}{2}\right) + \ln\left(\frac{2\Gamma(\frac{u+1}{2})^2}{u\Gamma(\frac{u}{2})^2}\right) \end{aligned}$ and Stirling's approximation for the Gamma function tells us that this ratio of Gamma functions tends to $1$ as $u \to \infty$ , so we deduce that $I \; = \; \ln\left(\tfrac{\pi}{2}\right)$ making the answer $\boxed{2\pi}$ .

Note that $\int_0^1 x^y dy=\frac{x-1}{\ln x}$ using this result our integral becomes $\begin{aligned} I &=\int_0^1\frac{x-1}{(x+1) \ln x}=\int_0^1\left(\frac{1}{x+1}\int_0^1x^ydy\right)dx \\&=\int_0^1\int_0^1\frac{x^y}{1+x}dy dx=\int_0^1\underbrace{\int_0^1\frac{x^y}{1+x}dx }_{I_1}dy \end{aligned}$ Since the integral $I_1=\int_0^1\frac{x^y}{1+x}dx =\frac{1}{2}\left(H_{\frac{ y}{2}}-H_{\frac{y-1}{2}}\right)=\frac{1}{2}\left(\psi^0\left(\frac{2y+1}{2}\right)-\psi^0\left(\frac{y+1}{2}\right)\right)$ is well know result which I have proved here by polynomial long division. Integrating $I_1$ we yield $\begin{aligned}I &= \int_0^1 I_1 dy =\left[\ln\Gamma\left(\frac{y+2}{2}\right)-\ln\Gamma\left(\frac{y+1}{2}\right)\right]_0^1\\& =\ln\left( \Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\right)=\ln\left(\frac{\sqrt{\pi}}{2}\sqrt{\pi}\right)\cdots(1)\\&=\ln\left(\frac{\pi}{2}\right)\end{aligned}$ we use the half gamma $\displaystyle \Gamma\left(\frac{1}{2}+n\right)=\frac{(2n)!}{4^n n!}\sqrt{\pi}$ argument in $(1)$ or using functional equation of gamma function we can write $(1)$ as $\displaystyle=\ln\left(\frac{1}{2}\Gamma^2\left(\frac{1}{2}\right)\right)=\ln\left(\frac{\pi}{2}\right)$

Alternatively $\begin{aligned}I_1 & =\int_0^1\frac{x^y}{1+x}dx =\int_0^1x^y \left(\sum_{r=0}^{\infty} (-1)^r x^r\right)dx\\&=\sum_{r=0}^{\infty} \frac{(-1)^r}{y+r+1}=\Phi\left(-1,1,y+1\right)\cdots(3)\end{aligned}$ where $\Phi(z,s,\alpha )$ is Lerch Transcendent function using the equation 5 and 6 we obtain $I= \frac{1}{2}\left(\psi^0\left(\frac{2y+1}{2}\right)-\psi^0\left(\frac{y+1}{2}\right)\right)$ To prove tha relationship between $ (3)$ and final result we can directly use series formula of digamma function.

The above integral can be related as

$\int_0^{\infty} \frac{\operatorname{tanh}x}{e^{2x}} \frac{dx}{x}=\int_0^1\frac{x-1}{(1+x)\ln x} dx=-\ln\left(\frac{2}{\pi}\right)$

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Sharing some of my work much related to this problem here