The integral shown is:

$\int_0^2 \frac{dx}{\sqrt{|x - 1|}}$

This is a piecewise integral because the absolute value function creates two cases:

1. $x - 1 \geq 0$ when $x \geq 1$, which means $|x - 1| = x - 1$
2. $x - 1 < 0$ when $x < 1$, which means $|x - 1| = 1 - x$

So, we split the integral into two parts: $\int_0^1 \frac{dx}{\sqrt{1 - x}} + \int_1^2 \frac{dx}{\sqrt{x - 1}}$

Solving each part:

1. $\int_0^1 \frac{dx}{\sqrt{1 - x}}$

This is a standard integral and can be solved using the substitution $u = 1 - x$, which gives: $\int \frac{dx}{\sqrt{1 - x}} = 2\sqrt{1 - x}$ Evaluating from $0$ to $1$: $2\sqrt{1 - x} \Bigg|_0^1 = 2(1 - 0) = 2$

2. $\int_1^2 \frac{dx}{\sqrt{x - 1}}$

Similarly, using the substitution $v = x - 1$, we have: $\int \frac{dx}{\sqrt{x - 1}} = 2\sqrt{x - 1}$ Evaluating from $1$ to $2$: $2\sqrt{x - 1} \Bigg|_1^2 = 2(1 - 0) = 2$

Final result:

Summing both parts, the total integral is: $2 + 2 = 4$

So, the value of the integral is $4$.

Would you like further details on any step or have any questions?

Here are five related questions to explore this concept:

1. How do you solve integrals involving absolute value functions in general?
2. Can you explain the substitution method used in these integrals?
3. What happens if the limits of integration were different?
4. How would the solution change if the exponent on the absolute value function was different (e.g., $|x - 1|^{3/2}$)?
5. Can you apply this method to integrals with discontinuities in the integrand?

Tip: When working with absolute value functions, always check if you need to break the integral into separate intervals where the absolute value changes its definition.