The given integral is:

$\int_0^1 \frac{\cosh^{-1}(\sqrt{x})}{\sqrt{x^2 - x}} \, dx.$

Let us proceed to evaluate it step-by-step.

Step 1: Simplify the integrand

The inverse hyperbolic cosine function, $\cosh^{-1}(z)$, satisfies the identity:

$\cosh^{-1}(z) = \ln(z + \sqrt{z^2 - 1}),$

for $z \geq 1$. Here, $\sqrt{x}$ is the argument, and it is valid since $x \in [0, 1]$.

Also, the term $\sqrt{x^2 - x}$ can be factored as:

$\sqrt{x^2 - x} = \sqrt{x(x - 1)}.$

Step 2: Variable substitution

To simplify the limits and integrand, substitute $x = t^2$, where $t \in [0, 1]$. Then:

$dx = 2t \, dt \quad \text{and} \quad \sqrt{x} = t.$

The integral becomes:

$\int_0^1 \frac{\cosh^{-1}(t)}{\sqrt{t^4 - t^2}} \cdot 2t \, dt.$

Step 3: Simplify further

Factor the denominator:

$\sqrt{t^4 - t^2} = \sqrt{t^2(t^2 - 1)} = t \sqrt{t^2 - 1}.$

Substitute this back:

$\int_0^1 \frac{\cosh^{-1}(t)}{t \sqrt{t^2 - 1}} \cdot 2t \, dt = 2 \int_0^1 \frac{\cosh^{-1}(t)}{\sqrt{t^2 - 1}} \, dt.$

Step 4: Further substitution for standard integral

Let $t = \cosh(u)$, which implies $dt = \sinh(u) \, du$, and:

$\sqrt{t^2 - 1} = \sqrt{\cosh^2(u) - 1} = \sinh(u).$

The integral becomes:

$2 \int_0^1 \frac{\cosh^{-1}(\cosh(u))}{\sinh(u)} \cdot \sinh(u) \, du = 2 \int_0^1 u \, du.$

Step 5: Evaluate the integral

The final integral is straightforward:

$2 \int_0^1 u \, du = 2 \left[ \frac{u^2}{2} \right]_0^1 = 2 \cdot \frac{1}{2} = 1.$

Final Answer:

$\boxed{1}.$

Would you like further clarification or details on any step?

Related Questions:

1. What is the definition and domain of the inverse hyperbolic cosine function?
2. How does the substitution $x = t^2$ simplify square root terms in integrals?
3. What are some other common integrals involving inverse hyperbolic functions?
4. Why is the substitution $t = \cosh(u)$ useful for this integral?
5. How do we verify the limits when making substitutions?

Tip: When dealing with inverse hyperbolic functions, their logarithmic representation can simplify evaluations significantly.