Answer
$\dfrac{3}{4}$
Work Step by Step
As we are given that $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt$
Now, consider $\ln t= a$ and $da=\dfrac{dt}{t}$
This implies: $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt=\int^{\ln 2}_{0} ( \cosh a) da=[ \sinh h a]^{\ln 2}_{0}$
and $= \sinh (\ln 2) -0$
As we know that $\sinh \theta=(\dfrac{e^{\theta} -e^{-\theta}}{2})$
Thus, we have, $ \sinh (\ln 2)= \dfrac{e^{\ln 2} -e^{-\ln 2}}{2}$
Hence, $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt=\dfrac{3}{4}$
