Evaluate the following integrals:

Let $(2-x)=t$, then $d t=-d x$, or $d x=-d t$

Hence, $\int \frac{1}{\sqrt{(2-x)^{2}+1}} d x=\int \frac{1}{t^{2}+1}(-d t)$

$\left.=-\int \frac{1}{t^{2}+1^{2}} d t=-\log \int\left(t+\sqrt{t^{2}+1}\right)\right)+c\left\{\right.$ since $\left.\left.\int \frac{1}{\sqrt{\left(a^{2}+x^{2}\right)}} d x=\log \left[x+\sqrt{\left(a^{2}\right.}+x^{2}\right)+c\right\}\right\}$

Put $t=2-x$

$\left.=-\log \int\left((2-\mathrm{x})+\sqrt{(2-\mathrm{x})^{2}+1}\right)\right)+\mathrm{c}$