Set $$ I_n :=\int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n \:\mathrm{d}x \qquad n=1,2,3,\cdots. $$

We have $$I_1 =\gamma, \quad I_2 =\log (2 \pi) - \frac 32, \quad I_3 = 6 \log A - \frac{31}{24}, \quad I_4 = 2 \log A + \frac{5 \zeta(3)}{2\pi^{2}}- \frac{49}{72}, \quad ...$$ where $A$ is the Glaisher-Kinkelin constant defined by $$ \begin{equation} \displaystyle A :=\lim_{n\to\infty}\frac{1^22^2\cdots n^n}{e^{-n^2/4}n^{\frac{n^2+n}{2}+\frac{1}{12}}}=1.28242712\cdots. \end{equation} $$

I wonder if there is a "simple" equivalent for $I_n$ as $n$ tends to $+\infty$?

Edit. I had designed the following integral $$\displaystyle \int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^2 \mathrm{d}x$$ which I submitted to American Mathematical Monthly (March 2012, problem 11629), the problem was then spread and came in this forum with different interesting solutions (I). An interesting general formula for $I_n$ has been found (II), but I don't think the latter formula is tractable for the above question on asymptotics.

Set $$ I_n :=\int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n \:\mathrm{d}x \qquad n=1,2,3,\cdots. $$

We have $$I_1 =\gamma, \quad I_2 =\log (2 \pi) - \frac 32, \quad I_3 = 6 \log A - \frac{31}{24}, \quad I_4 = 2 \log A + \frac{5 \zeta(3)}{2\pi^{2}}- \frac{49}{72}, \quad ...$$ where $A$ is the Glaisher-Kinkelin constant defined by $$ \begin{equation} \displaystyle A :=\lim_{n\to\infty}\frac{1^22^2\cdots n^n}{e^{-n^2/4}n^{\frac{n^2+n}{2}+\frac{1}{12}}}=1.28242712\cdots. \end{equation} $$

I wonder if there is a "simple" equivalent for $I_n$ as $n$ tends to $+\infty$?

Edit. I had designed the following integral $$\displaystyle \int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^2 \mathrm{d}x$$ which I submitted to American Mathematical Monthly (March 2012, problem 11629), the problem was then spread and came in this forum with different interesting solutions (I). An interesting general formula for $I_n$ has been found (II), but I don't think the latter formula is tractable for the above question on asymptotics.

Set $$ I_n :=\int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n \:\mathrm{d}x \qquad n=1,2,3,.... $$

We have $$I_1 =\gamma, \quad I_2 =\log (2 \pi) - \frac 32, \quad I_3 = 6 \log A - \frac{31}{24}, \quad I_4 = 2 \log A + \frac{5 \zeta(3)}{2\pi^{2}}- \frac{49}{72}, \quad ...$$ where $A$ is the Glaisher-Kinkelin constant defined by $$ \begin{equation} \displaystyle A :=\lim_{n\to\infty}\frac{1^22^2\cdots n^n}{e^{-n^2/4}n^{\frac{n^2+n}{2}+\frac{1}{12}}}=1.28242712\cdots. \end{equation} $$

I wonder if there is a "simple" equivalent for $I_n$ as $n$ tends to $+\infty$?

EDIT: Some complementary information. I've invented the pretty integral $$\displaystyle \int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^2 \mathrm{d}x$$ which I submitted to American Mathematical Monthly (March 2012, problem 11629), the problem was then spread and came in this forum with different interesting solutions (I). An interesting general formula for $I_n$ has been found (II), is it tractable for the above question on asymptotics? I call these integrals 'Binet-like integrals'. Others are related (IIIa), (IIIb) but I don't think these last can help for our specific question on asymptotics.

Set $$ I_n :=\int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n \:\mathrm{d}x \qquad n=1,2,3,\cdots. $$

We have $$I_1 =\gamma, \quad I_2 =\log (2 \pi) - \frac 32, \quad I_3 = 6 \log A - \frac{31}{24}, \quad I_4 = 2 \log A + \frac{5 \zeta(3)}{2\pi^{2}}- \frac{49}{72}, \quad ...$$ where $A$ is the Glaisher-Kinkelin constant defined by $$ \begin{equation} \displaystyle A :=\lim_{n\to\infty}\frac{1^22^2\cdots n^n}{e^{-n^2/4}n^{\frac{n^2+n}{2}+\frac{1}{12}}}=1.28242712\cdots. \end{equation} $$

I wonder if there is a "simple" equivalent for $I_n$ as $n$ tends to $+\infty$?

Edit. I had designed the following integral $$\displaystyle \int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^2 \mathrm{d}x$$ which I submitted to American Mathematical Monthly (March 2012, problem 11629), the problem was then spread and came in this forum with different interesting solutions (I). An interesting general formula for $I_n$ has been found (II), but I don't think the latter formula is tractable for the above question on asymptotics.