1. Set up integration by parts by choosing $u$ and $\frac{\text{d}v}{\text{d}x}$.

Let $u = \ln(x^2 - 4)$ and $\frac{\text{d}v}{\text{d}x} = x$.
Then $\frac{\text{d}u}{\text{d}x} = \frac{2x}{x^2 - 4}$ and $v = \frac{1}{2}x^2$.

2. Apply the integration by parts formula.

$\begin{aligned} \int x \ln(x^2 - 4) \text{ d}x &= \frac{1}{2}x^2 \ln(x^2 - 4) - \int \frac{1}{2}x^2 \left( \frac{2x}{x^2 - 4} \right) \text{ d}x \\ &= \frac{1}{2}x^2 \ln(x^2 - 4) - \int \frac{x^3}{x^2 - 4} \text{ d}x \end{aligned}$

3. Rewrite the remaining integral using algebraic manipulation.

By algebraic manipulation or long division, we can rewrite the improper fraction:
$\frac{x^3}{x^2 - 4} = \frac{x(x^2 - 4) + 4x}{x^2 - 4} = x + \frac{4x}{x^2 - 4}$

4. Integrate the remaining terms and simplify to the final answer.

$\begin{aligned} \int x \ln(x^2 - 4) \text{ d}x &= \frac{1}{2}x^2 \ln(x^2 - 4) - \int \left( x + \frac{4x}{x^2 - 4} \right) \text{ d}x \\ &= \frac{1}{2}x^2 \ln(x^2 - 4) - \left( \frac{1}{2}x^2 + 2\ln(x^2 - 4) \right) + C \\ &= \left(\frac{1}{2}x^2 - 2\right)\ln(x^2 - 4) - \frac{1}{2}x^2 + C \\ &= \frac{1}{2}(x^2 - 4)\ln(x^2 - 4) - \frac{1}{2}x^2 + C \end{aligned}$

1. Find the differential $dx$ in terms of $du$.

Given $x = a \cos u$, differentiating with respect to $u$ gives:
$\frac{dx}{du} = -a \sin u$
$\implies dx = -a \sin u \text{ d}u$

2. Determine the new limits of integration.

When $x = 0$:
$a \cos u = 0 \implies \cos u = 0 \implies u = \frac{\pi}{2}$

When $x = a$:
$a \cos u = a \implies \cos u = 1 \implies u = 0$

3. Substitute into the integral and simplify the integrand.

$\int_{0}^{a} \sqrt{a^2 - x^2} \text{ d}x$
$= \int_{\frac{\pi}{2}}^{0} \sqrt{a^2 - (a \cos u)^2} (-a \sin u) \text{ d}u$
$= \int_{\frac{\pi}{2}}^{0} \sqrt{a^2(1 - \cos^2 u)} (-a \sin u) \text{ d}u$
$= \int_{\frac{\pi}{2}}^{0} \sqrt{a^2 \sin^2 u} (-a \sin u) \text{ d}u$
Since $u \in [0, \frac{\pi}{2}]$, $\sin u \ge 0$, so $\sqrt{a^2 \sin^2 u} = a \sin u$:
$= \int_{\frac{\pi}{2}}^{0} (a \sin u)(-a \sin u) \text{ d}u$
$= \int_{\frac{\pi}{2}}^{0} -a^2 \sin^2 u \text{ d}u$
$= a^2 \int_{0}^{\frac{\pi}{2}} \sin^2 u \text{ d}u$

4. Integrate using the double angle identity.

Using the identity $\sin^2 u = \frac{1 - \cos 2u}{2}$:
$a^2 \int_{0}^{\frac{\pi}{2}} \sin^2 u \text{ d}u = a^2 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2u}{2} \text{ d}u$
$= \frac{a^2}{2} \left[ u - \frac{1}{2} \sin 2u \right]_{0}^{\frac{\pi}{2}}$

5. Evaluate the definite integral to find the exact value.

$= \frac{a^2}{2} \left( \left( \frac{\pi}{2} - \frac{1}{2} \sin \pi \right) - \left( 0 - \frac{1}{2} \sin 0 \right) \right)$
$= \frac{a^2}{2} \left( \frac{\pi}{2} - 0 - 0 \right)$
$= \frac{\pi a^2}{4}$