We have the following integral: \begin{align} I_n=\int_0^1 P_n(x)\frac{\ln x}{1-x}dx = \frac{a_n\zeta(2)+b_n}{d_n^2} \end{align} $P_n(x)$ is a polynomial with integer coefficients and degree $n$. $a_n,b_n$ are integers and $d_n$ is the less common multiple of all consecutive natural numbers up to $n$, so, $d_n^2=$ LCM$(1,2,...,n)^2$.
Since, \begin{align} \int_0^1 \frac{1}{1-ay}dy=-\frac{\ln (1-a)}{a} \end{align}
we can use $a=1-x$ to have \begin{align} I_n=\int_0^1 P_n(x)\frac{\ln x}{1-x}dx &= -\int_0^1\int_0^1 P_n(x)\frac{1}{1-(1-x)y}dxdy\\ &=\frac{a_n\zeta(2)+b_n}{d_n^2} \end{align}
now it's easier to integrate $I_n$ by parts $n$ times.
Is there a polynomial $P_n(x)$ with integer coefficients and degree $n$ such that $I_n d_n^2 \to 0$ as $n \to \infty\:?$
I'm using this estimation: $d_n^2<e^{2An}$ for some real $A>1$.