If $b = 0$, both sides of the equation are zero. So assume $|a| > |b| > 0$. By rescaling, the integral can be written $\int_ 0^\infty I(u; x)\, \frac{dx}{b}$ where
$\displaystyle I(u; x) = \frac{\sinh{x}}{\cosh{u x} + \cosh{x}}$
and $u = \frac{a}{b}$. Since the hyperbolic cosine is even, the value of the integral is unchanged when $u$ is replaced by $-u$. So we may assume, without loss of generality, that $u > 1$. Using the identity
$\displaystyle \cosh{ux} + \cosh{x} = \frac{e^{-ux}}{2}(e^{(u+1) x} + 1)(e^{(u-1) x} + 1)$
we obtain
$\displaystyle I(u; x) = \frac{e^{(u+1)x} - e^{(u-1) x}}{(e^{(u+1)x} + 1)(e^{(u-1) x} + 1)} = \frac{1}{e^{(u-1) x} + 1} - \frac{1}{e^{(u+1) x} + 1}$.
Since $u - 1$ and $ u + 1$ are positive, the last expression can written as the series
$\displaystyle \sum_{n = 1}^\infty (-1)^{n-1} (e^{-n (u-1) x} - e^{-n (u+1) x})$.
For $T > 0$,
$\displaystyle \sum_ {n = 1}^\infty (-1)^{n-1} \int_ T^\infty (e^{-n (u-1) x} - e^{-n (u+1)})\, dx$
$\displaystyle = \sum_{n = 1}^\infty \frac{(-1)^{n-1}{n}\left(\frac{e^{-(u-1) x}{u-1} - \frac{e^{-(u+1) x}{u+1}\right)$
$\displaystyle = \frac{\ln(1+e^{-(u-1) T})}{u-1} - \frac{\ln(1+e^{-(u+1)T})}{u+1}$,
which tends to zero as $T \to \infty$. Thus
$\displaystyle \int_0^\infty I(u; x)\, \frac{dx}{b} = \sum_{n = 1}^\infty (-1)^{n-1} \int_ 0^\infty (e^{-n (u-1)x} - e^{-n (u+1) x})\, \frac{dx}{b}\quad (*)$.
Since the integrals in (*) converge to $\frac{1}{n}\left(\frac{1}{u-1} - \frac{1}{u+1}\right)$, we deduce
$\displaystyle \int_ 0^\infty I(u; x)\, \frac{dx}{b} = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n}\left(\frac{1}{u-1} - \frac{1}{u+1}\right) = \frac{2 \ln{2}}{b (u^2-1)} = \frac{2b\ln{2}}{a^2-b^2}$.
