An integral Thing I guess

Making the substitutions $u = \sqrt{x^2+1} + x$ and then $v = u^{-1}$ we note that $\begin{aligned} \int_\mathbb{R} \left(\frac{\sinh^{-1}x}{x}\right)^2\,dx & = \; 2\int_{\mathbb{R}} \left(\frac{\ln(\sqrt{x^2+1}+x)}{x}\right)^2\,dx \; = \; 4\int_1^\infty \frac{u^2+1}{(u^2-1)^2}(\ln u)^2\,du \\ & = \; 4\int_0^1 \frac{v^2+1}{(1-v^2)^2}(\ln v)^2\,dv \; = \; 4\int_0^1 \left(\sum_{n \ge 0} (2n+1)v^{2n}\right)(\ln v)^2\,dv \\ & = \; 8\sum_{n\ge 0} \frac{1}{(2n+1)^2} \;= \; 8 \times \tfrac34\zeta(2) \; = \; \pi^2 \end{aligned}$ using the fact that $\int_0^1 v^m (\ln v)^2\,dv \; = \; \frac{2}{(m+1)^3} \hspace{2cm} m \ge 0$ Thus the answer is $\boxed{2}$ .

integral((sinh^(-1)(x))/x)^2 dx = 2 Li 2(-e^(-sinh^(-1)(x))) - 2 Li 2(e^(-sinh^(-1)(x))) - sinh^(-1)(x) ((sinh^(-1)(x))/x - 2 log(1 - e^(-sinh^(-1)(x))) + 2 log(e^(-sinh^(-1)(x)) + 1)) + constant