Find the integrals

Note that we have the expansion of the so-called "rising factorial" $x^{(n)} \; =\; \prod_{j=0}^{n -1}(x+j) \; = \; \sum_{j=1}^n \left[{n \atop j}\right] x^j \hspace{2cm} n \ge 1$ where $\left[{n \atop k}\right]$ is an unsigned Stirling number of the first kind. Also note that $(1 - x)^{-k-1} \; = \; \sum_{r=0}^\infty \frac{(r+1)^{(k)}}{k!} x^r \hspace{2cm} |x| < 1\,,\, k \ge 1$ while $\int_0^1 x^u (\ln x)^v\,dx \; = \; (-1)^{v+1} \frac{v!}{(u+1)^{v+1}} \hspace{2cm}u,v \in \mathbb{N} \cup \{0\}$ Thus $\begin{aligned} I_k & = \; \int_0^1 \left(\frac{\ln x}{1-x}\right)^{k+1}\,dx \; = \; \sum_{r=0}^\infty \frac{(r+1)^{(k)}}{k!} \int_0^1 x^r (\ln x)^{k+1}\,dx \\ & = \; (-1)^{k+1}\sum_{r=0}^\infty \frac{(r+1)^{(k)}}{k!} \times \frac{(k+1)!}{(r+1)^{k+2}} \; = \; (-1)^{k+1}(k+1)\sum_{r=0}^\infty \frac{(r+1)^{(k)}}{(r+1)^{k+2}} \\ & = \; (-1)^{k+1}(k+1)\sum_{r=0}^\infty \frac{1}{(r+1)^{k+2}} \sum_{j=1}^k \left[{k \atop j}\right] (r+1)^j \; = \; (-1)^{k+1}(k+1)\sum_{j=1}^k \left[ {k \atop j} \right]\zeta(k+2-j) \end{aligned}$ Thus $\begin{aligned} I_2 & = \; -3\left[\zeta(3) + \zeta(2)\right] \; = \; -3\zeta(3) - \tfrac12\pi^2 \\ I_3 & = \; 4\left[\zeta(4) + 3\zeta(3) + \zeta(2)\right] \; = \; \tfrac{4}{45}\pi^4 + 12\zeta(3) + \tfrac23\pi^2 \end{aligned}$ and so the desired quantity is $I_3 + 4I_2 \; = \; \tfrac{4}{45}\pi^4 - \tfrac43\pi^2 \; =\; \frac{4\pi^2(\pi^2 - 15)}{45}$ making the answer $4 + 2 + 45 + 2 + 15 = \boxed{68}$ .