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

Put $t=2-x$

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