Integrate the function: $\frac{1}{\sqrt{(2-x)^{2}+1}}$

(A) Let $2-x=t$.
Then,$-dx = dt$,which implies $dx = -dt$.
Substituting these into the integral:
$\int \frac{1}{\sqrt{(2-x)^{2}+1}} dx = -\int \frac{1}{\sqrt{t^{2}+1}} dt$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^{2}+a^{2}}} dx = \log |x + \sqrt{x^{2}+a^{2}}| + C$,we get:
$= -\log |t + \sqrt{t^{2}+1}| + C$.
Substituting $t = 2-x$ back into the expression:
$= -\log |2-x + \sqrt{(2-x)^{2}+1}| + C$.
Since $-\log|u| = \log|1/u|$,this can be written as:
$= \log \left| \frac{1}{(2-x) + \sqrt{x^{2}-4x+5}} \right| + C$,where $C$ is an arbitrary constant.