= Solution
For $x>0$, the definition of <hyperbolic cosine> simplifies the composite function exactly:
$$
\cosh(\ln x)=\frac{e^{\ln x}+e^{-\ln x}}2=\frac12\left(x+\frac1x\right).
$$
Let $h=x-2$. The <geometric series> gives
$$
\frac1{2+h}=\frac12\left(1-\frac h2+\frac{h^2}4-\frac{h^3}8+O(h^4)\right).
$$
Adding $2+h$ and dividing by two gives the requested <Taylor expansion>:
$$
\boxed{\cosh(\ln x)=\frac54+\frac38(x-2)+\frac1{16}(x-2)^2-\frac1{32}(x-2)^3+O((x-2)^4).}
$$
The underlying infinite geometric expansion has $|x-2|<2$.
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