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Thanks to Gary Liang and metamorphy who had given me links of relevant materials so that the closed form of our integral can be found as
$$ \boxed{\int_0^1 \frac{x^{n+1}}{(1-x) \ln ^{n+1}(1-x)} d x =-\frac{n+1}{n !} \sum_{k=1}^n\left(\begin{array}{l} n \\ k \end{array}\right)(-1)^{n-k}(k+1)^{n-1} \ln (k+1)} $$ By this formula, we can evaluate the integral by finding the series instead of evaluating multiple integrals.
Couple of days ago, I encountered the integral $$ J_2=\int_0^1 \frac{x^2}{(1-x) \ln ^2(1-x)} d x. $$ For convenience, I first transformed the integral by the substitution $x\mapsto 1-x$ and then made use of double integral.
$$ \begin{aligned} J_2 &=\int_0^1 \frac{(1-x)^2}{x \ln ^2x} d x =-\int_0^1(1-x)^2 d\left(\frac{1}{\ln x}\right) \\&=2 \int_0^1 \frac{x-1}{\ln x} d x=2 \int_0^1 \int_0^1 x^t d t d x\\&=2 \int_0^1 \int_0^1 x^t d x d t= 2 \int_0^1\left[\frac{x^{t+1}}{t+1}\right]_0^1 d t\\&= 2 \int_0^1 \frac{1}{t+1} d t= 2 \ln 2\\ \end{aligned} $$
Then I tried generalise $J_2$ by raising the power by $2$ to $n+1$ and used similar technique to get
$$J_{n+1}=\int_0^1 \frac{x^{n+1}}{(1-x) \ln ^{n+1}(1-x)} d x= -\frac{n+1}{n}\int_0^1 \left(\frac{1-x}{\ln x}\right)^{n} d x $$
Replacing $\phi$ by $1$ in my post, we have $$ \begin{aligned} J_{n+1}&= (-1)^{n+1}\frac{n+1}{n}\int_0^1\left(\frac{x-1}{\ln x}\right)^n d x \\ & = (-1)^{n+1}\frac{n+1}{n}\int_0^1 \left(\underbrace{\int_0^1 \int_0^1 \cdots \int_0^1}_{n \text { integral signs }} x^{t_1+t_2+\ldots+t_n} d t_1 d t_2 \cdots d t_n\right) dx \\&= (-1)^{n+1}\frac{n+1}{n} \underbrace{\int_0^1\int_0^1\cdots \int_0^1}_{n \text { integral signs } }\left(\int_0^1 x^{t_1+t_2+\ldots+t_n} d x\right) d t_1 d t_2 \cdots d t_n \\ \int_0^1 \frac{x^{n+1}}{(1-x) \ln ^{n+1}(1-x)} d x&= (-1)^{n+1}\frac{n+1}{n}\int_0^1\int_0^1 \cdots \int_0^1\frac{1}{1+t_1+t_2+\cdots+t_n} d t_1 d t_2 \cdots d t_n \blacksquare\\ \end{aligned} $$
For confirmation, we start with $$J_2=2\int_0^1 \frac{1}{1+t_1} d t_1=2\ln 2$$
$$ \begin{aligned} J_3&=-\frac{3}{2} \int_0^1 \int_0^1 \frac{1}{1+t_1+t_2} d t_1 d t_2 \\&=-\frac{3}{2} \int_0^1\left[\ln \left(1+t_1+t_2\right)\right]_0^1 d t_2 \\ &=-\frac{3}{2} \int_0^1\left[\ln \left(2+t_2\right)-\ln \left(1+t_2\right)\right] d t_2 \\ &=-\frac{3}{2}\left[\left(2+t_2\right)\left(\ln \left(2+t_2\right)-1\right) -\left(1+t_2\right)\left(\ln \left(1+t_2\right)-1\right)\right]\,_0^1\\ &=\frac{3}{2} \ln \left(\frac{16}{27}\right) \end{aligned} $$
and so on. We can find $J_{n}$ as long as we could repeatedly find $\int_0^1x^k\ln x dx$.
My question is how to simplify the multiple integral or obtain any other closed form. Your comments and alternative methods are highly appreciated.