Integral med

Here's the idea:

• Use the substitution $u = \ln x$ to rewrite the integral as $\frac{1}{2}\int_{-\infty}^0 \frac{u^2}{\cosh u}\,du$
• Note that this integrand is even, so the integral equals $\frac{1}{4} \int_{-\infty}^\infty \frac{u^2}{\cosh u}\,du$
• Compare this to the contour integral $\oint_\gamma \frac{z^2}{\cosh z}\,dz$ where $\gamma$ goes along the lines connecting the points, in order, $-R, R, R+\pi i, -R+\pi i, -R.$ Since this encloses a simple pole at $z=\frac{\pi}{2}i$ (and no other singularities), this shows $\int_{-\infty}^\infty \frac{2u^2 - \pi^2}{\cosh u}\,du = 2\pi i \cdot \frac{\left(\pi i / 2\right)^2}{\sinh\left(\pi i / 2\right)} = -\frac{\pi^3}{2}$ or by solving for the integral we want, $\frac{1}{4}\int_{-\infty}^\infty \frac{u^2}{\cosh u}\,du = -\frac{\pi^3}{16} + \frac{\pi^2}{8}\int_{-\infty}^\infty \frac{1}{\cosh u}\,du$
• Use the substitution $w = e^u$ to show $\int_{-\infty}^\infty \frac{1}{\cosh u}\,du = \int_0^\infty \frac{2}{1 + w^2}\,dw = 2\arctan w \Big|_0^\infty = \pi$

Put it all together: $\int_0^1 \frac{\ln^2(x)}{1+x^2}\,dx = \frac{1}{4}\int_{-\infty}^\infty \frac{u^2}{\cosh u}\,du = -\frac{\pi^3}{16} + \frac{\pi^2}{8}(\pi) = \frac{\pi^3}{16}$ so that $m=3, n=4 \implies m\cdot n = \boxed{12}$