Wish $e^{-2 x}$ wasn't there

If we consider $\begin{array}{rcl} \displaystyle \int_0^\infty \frac{\ln(1-b + be^{-2x})}{\sinh x}\,dx & = & \displaystyle 2\int_0^\infty \frac{\ln(1-b + be^{-2x})}{e^x-e^{-x}}\,dx \\ & = & \displaystyle 2\int_0^\infty \frac{\ln(1 - b + be^{-2x})}{1 - e^{-2x}}\,e^{-x}\,dx \\ & = & \displaystyle 2\int_0^1 \frac{\ln(1 - b(1-u^2))}{1 - u^2}\,du \\ & = & \displaystyle -2\sum_{n=1}^\infty \frac{b^n}{n} \int_0^1 (1-u^2)^{n-1}\,du \\ & = & \displaystyle -2\sum_{n=1}^\infty \frac{b^n}{n} \int_0^{\frac12\pi} \cos^{2n-1}\theta\,d\theta \\ & = & \displaystyle -2\sum_{n=1}^\infty \frac{2^n b^n (n-1)!}{n(2n-1)!!} \\ & = & -2\left(\sin^{-1}\sqrt{b}\right)^2 \end{array}$ for any $0 < b < 1$ , using the substitutions $u = e^{-x}$ and $u = \sin\theta$ , and a standard formula for $\big(\sin^{-1}\theta\big)^2$ , we deduce that $f(t) \; = \; -2\big(\sin^{-1}|\sin t|\big)^2\;,$ which makes $f(7) = -2(7-2\pi)^2 = \boxed{-1.02765}$ .

Use Feynmann's differentiation under the integral sign to differentiate it wrt t . You would get that f'(t) = -4arctan(tan(t)) . As f(0) = 0 . We integrate f'(t) from 0 to 7. As it is a discontinuous function it would be the same as the -4 times area of a right angled isosceles triangle of base and height (2 $\pi$ -7)
[See the graph for clarity] Which gives the result as -1.207