This question is from the MIT Integration Bee 2023 Semifinal 2 and is Question 3. The goal is to show $$\int_{0}^{\infty} \frac{\tanh(x)}{x\cosh(2x)}\,\textrm{d}x = \log(2)$$
Ideally, this should be done in three minutes or less. Something I tried doing at first was Feynman's trick, but I ran into a cancellation when trying to evaluate $$I^\prime(t) = \int_{0}^{\infty} \operatorname{sech}^2(tx)\operatorname{sech}(2x)\,\textrm{d}x$$ in that I'd get something along the lines of $I^\prime(t) = -I^\prime(t)$ in the end. Is my approach correct, or is there something that allows me to ignore the hyperbolic functions?
Note $$ \frac{\tanh(x)}{\cosh(2x)}=\frac2{1+e^{2x}}-\frac2{1+e^{4x}}. $$ Let $$ f(x)=\frac2{1+e^{x}} $$ and then $$\int_{0}^{\infty} \frac{\tanh(x)}{x\cosh(2x)}\,\textrm{d}x =\int_0^\infty\frac{f(2x)-f(4x)}x\,\textrm{d}x=(f(\infty)-f(0))\ln(\frac24)=\ln2 $$ by Frullani's Integral