Anothe Integral Thing I guess

Note that $\int_0^1 x^m (\ln x)^n \; = \; \frac{(-1)^n n!}{(m+1)^{n+1}} \hspace{2cm} m,n \in \mathbb{N} \cup \{0\}$ and that we can find rationals $\alpha_{k,j}$ for integers $0 \le j \le k$ such that ${n+k \choose n} \; \equiv \; \sum_{j=0}^k \alpha_{k,j}(2n+1)^j$ Then $\begin{aligned} X_K & = \; \int_0^1 \frac{(\ln x)^6}{(1-x^2)^K}\,dx \;= \; \int_0^1 \sum_{n \ge 0} {n+K-1 \choose n} x^{2n} (\ln x)^6\,dx \\ & = \; \sum_{n \ge 0} {n+K-1 \choose n} \frac{720}{(2n+1)^7}\; = \; 720\sum_{n \ge 0} \sum_{j=0}^{K-1} \alpha_{K-1,j} \frac{1}{(2n+1)^{7-j}} \\ & = \; 720 \sum_{j=0}^{K-1} \alpha_{K-1,j}\left(1 - \frac{1}{2^{7-j}}\right)\zeta(7-j) \end{aligned}$ for any imteger $3 \le K \le 6$ .

Making the substitutions $u = \sqrt{x^2+1} + x$ and then $v = u^{-1}$ we note that $\begin{aligned} \int_\mathbb{R} \left(\frac{\sinh^{-1}x}{x}\right)^6\,dx & = \;2 \int_{0}^{\infty} \left(\frac{\ln(\sqrt{x^2+1}+x)}{x}\right)^6\,dx \; = \; 64\int_1^\infty \frac{u^4(u^2+1)}{(u^2-1)^6}(\ln u)^6\,du \\ & = \; 64\int_0^1 \frac{v^4(v^2+1)}{(1-v^2)^6}(\ln v)^6\,dv \; = \; 64\big(2X_6 - 5X_5 + 4X_4 - X_3\big) \\ & = \; 64 \times 720\left[ \begin{array}{l}(2\alpha_{5,0} - 5\alpha_{4,0} + 4\alpha_{3,0} - \alpha_{2,0})\tfrac{127}{128}\zeta(7) \\ + (2\alpha_{5,1} - 5\alpha_{4,1} + 4\alpha_{3,1} - \alpha_{2,1})\tfrac{63}{64}\zeta(6) \\ + (2\alpha_{5,2} - 5\alpha_{4,2} + 4\alpha_{3,2} - \alpha_{2,2})\tfrac{31}{32}\zeta(5)\\ + (2\alpha_{5,3} - 5\alpha_{4,3} + 4\alpha_{3,3})\tfrac{15}{16}\zeta(4) \\ + (2\alpha_{5,4} - 5\alpha_{4,4})\tfrac{7}{8}\zeta(3) + 2\alpha_{5,5}\tfrac34\zeta(2) \end{array}\right] \end{aligned}$ After a lot of calculation, this gives us $\int_\mathbb{R} \left(\frac{\sinh^{-1}x}{x}\right)^6\,dx \; = \; 3\pi^2 - \tfrac52\pi^4 + \tfrac{9}{40}\pi^6$ (the terms in $\zeta(3), \zeta(5), \zeta(7)$ all cancel out). Thus the answer is $3 + 5 + 2 + 9 + 40 = \boxed{59}$ .