Answer
The integral $\int_{1}^{\infty} \dfrac{\ln x}{\sinh x} \ dx$ converges.
Work Step by Step
We are given the function
$f(x)=\int_{1}^{\infty} \dfrac{\ln x}{\sinh x} \ dx$
Since, $e^{-x} \lt \dfrac{e^x}{2}$ and $\sinh x=\dfrac{e^x-e^{-x}}{2}$
This yields:
$\dfrac{\ln x}{\sinh x} \leq \dfrac{4x}{e^x} $
Consider the integral
$\int_{1}^{\infty} \dfrac{4x}{e^x} dx=[-4xe^{-x} ]_1^{\infty}+4 \int_1^{\infty} e^{-x} dx \\=\dfrac{4}{e}-4[0-e^{-1}]\\=\dfrac{4}{e}+\dfrac{4}{e}\\=\dfrac{8}{e}$
Thus, the integral $\int_{1}^{\infty} \dfrac{4x dx}{e^x}$ converges to $\dfrac{8}{e}$. Therefore, by the comparison test, the integral $\int_{1}^{\infty} \dfrac{\ln x}{\sinh x} \ dx$ converges as well.
