I am reading the paper A note on the Irrationality of ζ(2) and ζ(3) by F. Beukers. In equation $(7)$, we have $$I_n=\int_{(0,1)^3}\frac{x^n(1-x)^ny^n(1-y)^nw^n(1-w)^n}{(1-(1-xy)w)^{n+1}} dx dy dw \tag{1}$$
Using this above equation, I need to prove that there exists $A_n,B_n\in\mathbb{Z}$ such that $$I_n=\frac{A_n+B_n\zeta(3)}{d_n^3}\tag{2}$$ where $d_n= \operatorname{lcm} (1,2,...,n)$
I tried using $n$-fold partial integration by parts with respect to $w$ in $(1)$ by taking $u=w^n(1-w)^n$ and $v=\frac{x^n(1-x)^ny^n(1-y)^n}{(1-(1-xy)w)^{n+1}}$, so that $$I_n=\int_{(0,1)^3}\frac{x^n(1-x)^ny^n(1-y)^nP_n(w)}{(1-(1-xy)w)(1-xy)^n} dx dy dw$$ where $P_n(w)=\frac{d^n}{dw^n}(\frac{w^n(1-w)^n}{n!})$ is the shifted Legendre Polynomial.
Now $$P_n(w)=\sum_{i=0}^{n}(-1)^i \binom{n}{i}\binom{n+i}{i} w^i$$ $$(1-x)^n=\sum_{j=0}^{n}(-1)^j \binom{n}{j} x^j$$ $$(1-y)^n=\sum_{k=0}^{n}(-1)^k \binom{n}{k} y^k$$ $$\frac{1}{1-(1-xy)w}=\sum_{r=0}^{\infty}(1-xy)^r w^r$$ So $I_n$ becomes $$I_n=\sum_{i,j,k=0}^n(-1)^{i+j+k}\binom{n}{i}\binom{n+i}{i}\binom{n}{j}\binom{n}{k}\sum_{r=0}^{\infty}\int_{(0,1)^3} \frac{x^{n+j} y^{n+k} w^{i+r}(1-xy)^r}{(1-xy)^n}dx dy dw$$ Thank You.