Given the limit
$$\lim_{n \to \infty}{\ln n\over e^n}\int_{0}^{n}{\sinh{x}\over \ln(1+x)}\mathrm dx={1\over 2}\tag1$$
How did I came across this integral $(1)$?
I was observing this integral $(2)$ $$\int_{0}^{n}{\sinh{x}\over \ln(1+x)}\mathrm dx=C\tag2$$ I notice that $$C\approx e^n$$
With a few lucky guess, I found it to be $$\ln C\approx n-\ln(\ln n)-\ln2$$
and simplifies it to $(1)$
Does the limit in $(1)$ hold?
Note that
$$f(y)={\ln y\over e^y}\int_{0}^{y}{\sinh{x}\over \ln(1+x)}\mathrm dx=\frac{\int_{0}^{y}{\sinh{x}\over \ln(1+x)}\mathrm dx}{{e^y\over \ln y}}$$
by l'Hospital $y\to \infty$
$$\frac{\int_{0}^{y}{\sinh{x}\over \ln(1+x)}\mathrm dx}{{e^y\over \ln y}}\stackrel{H.R.}\implies \frac{\sinh{y}\over \ln(1+y)} {{e^y\ln y-\frac{e^y}{y}\over \ln^2 y}}\sim \frac{\frac{e^y}{2\ln y} }{{e^y\ln y-\frac{e^y}{y}\over \ln^2 y}}=\frac{e^y}{2\ln y}\frac{\ln^2y}{e^y(\ln y-\frac1y)}=\frac{1}{2}\frac{\ln y}{\ln y-\frac1y}\to \frac12$$