The judges claim that the closed form of $$\int_0^\infty\frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)}\text{d}x$$ is $\frac{\text{log}(3)}{2},$ although no internet calculator except Wolfram|Alpha finds a closed form. WA returns that result for this integral.
What steps must to be taken to obtain this result? What is the indefinite integral? Work from both contestants can be seen in this YouTube video.
On
This is a similar observation to @heropup's existing answer, but as @PedroTamaroff requested I'll note in an answer that, since$$\frac{e^{3y}-e^y}{(e^y+1)(e^{3y}+1)}=\frac{(e^{3y}+1)-(e^y+1)}{(e^y+1)(e^{3y}+1)}=\frac{1}{e^y+1}-\frac{1}{e^{3y}+1},$$this is a Frullani integral with $f(y):=\frac{1}{e^y+1},\,a:=1,\,b:=3$, so is $\frac12\ln3$ because $f(0)=\frac12,\,f(\infty)=0$.
Consider the generalized integrand for $a > 0$ $$f(x;a) = \frac{e^{ax} - e^x}{x(1 + e^{ax})(1 + e^x)}.$$ Then $$\frac{\partial f}{\partial a} = \frac{e^{ax}}{(1 + e^{ax})^2}.$$ Thus $$I(a) = \int_{x=0}^\infty f(x;a) \, dx$$ satisfies $$I'(a) = \int_{x=0}^\infty \frac{\partial f}{\partial a} \, dx = \left[ - \frac{1}{a(1 + e^{ax})} \right]_{x=0}^\infty = \frac{1}{2a},$$ so $$I(a) = \frac{\log a}{2} + C$$ for some constant $C$ satisfying $I(1) = 0$, since $f(x;1) = 0$. It follows that $C = 0$ and the desired integral is $$I(3) = \frac{\log 3}{2}.$$