Answer
$$4$$
Work Step by Step
\begin{align*}
\int_{0}^{2} \frac{d x}{\sqrt{|x-1|}}&=\int_{0}^{1} \frac{d x}{\sqrt{1-x}}+\int_{1}^{2} \frac{d x}{\sqrt{x-1}}\\
&=\lim _{b \rightarrow 1^{-}}\int_{0}^{b} \frac{d x}{\sqrt{1-x}}+\lim _{c \rightarrow 1^{+}}\int_{b}^{2} \frac{d x}{\sqrt{x-1}}\\
&=\lim _{b \rightarrow 1^{-}}[-2 \sqrt{1-x}]\bigg|_{0}^{b}+\lim _{c \rightarrow 1^{+}}[2 \sqrt{x-1}]\bigg|_{c}^{2}\\
&=\lim _{b \rightarrow 1^{-}}[(-2 \sqrt{1-b})-(-2 \sqrt{1-0})]+\lim _{c \rightarrow 1^{+}}[2 \sqrt{2-1}-(2 \sqrt{c-1})]\\
&=0+2+2-0=4
\end{align*}
