Without the use of a calculator, show that $\int_{-2}^{2} \frac{|x-2|}{x^2 - 2x + 4} \,\mathrm{d}x = p \ln 3 + q \pi ,$ where $p$ and $q$ are exact real constants to be determined.

1. Resolve the modulus based on the limits of integration.

For $x \in [-2, 2]$, $x - 2 \le 0$. Hence, $|x - 2| = -(x - 2) = 2 - x$.

The integral simplifies to:
$\int_{-2}^{2} \frac{2 - x}{x^2 - 2x + 4} \,\mathrm{d}x$

2. Split the integrand to use the derivative of the denominator.

The derivative of the denominator is $\frac{\mathrm{d}}{\mathrm{d}x}(x^2 - 2x + 4) = 2x - 2$.

Express the numerator $2 - x$ in terms of $2x - 2$:
$2 - x = -\frac{1}{2}(2x - 2) + 1$

Split the integral into two parts:
$\int_{-2}^{2} \left( -\frac{1}{2}\frac{2x - 2}{x^2 - 2x + 4} + \frac{1}{x^2 - 2x + 4} \right) \,\mathrm{d}x$

3. Evaluate the first integral using the logarithmic form.

Using the standard form $\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)} \,\mathrm{d}x = \ln |\mathrm{f}(x)|$:
$-\frac{1}{2} \int_{-2}^{2} \frac{2x - 2}{x^2 - 2x + 4} \,\mathrm{d}x = \left[ -\frac{1}{2} \ln(x^2 - 2x + 4) \right]_{-2}^{2}$
$= -\frac{1}{2} \left( \ln(2^2 - 2(2) + 4) - \ln((-2)^2 - 2(-2) + 4) \right)$
$= -\frac{1}{2} (\ln 4 - \ln 12)$
$= -\frac{1}{2} \ln\left(\frac{4}{12}\right) = -\frac{1}{2} \ln\left(\frac{1}{3}\right) = \frac{1}{2} \ln 3$

4. Complete the square for the denominator of the second integral.

For the second integral, complete the square in the denominator:
$x^2 - 2x + 4 = (x - 1)^2 - 1^2 + 4 = (x - 1)^2 + 3$

The second integral becomes:
$\int_{-2}^{2} \frac{1}{(x - 1)^2 + (\sqrt{3})^2} \,\mathrm{d}x$

5. Evaluate the second integral using the inverse tangent form.

Using the standard integral $\int \frac{1}{x^2 + a^2} \,\mathrm{d}x = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$:
$\int_{-2}^{2} \frac{1}{(x - 1)^2 + (\sqrt{3})^2} \,\mathrm{d}x = \left[ \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x - 1}{\sqrt{3}}\right) \right]_{-2}^{2}$
$= \frac{1}{\sqrt{3}} \left( \tan^{-1}\left(\frac{2 - 1}{\sqrt{3}}\right) - \tan^{-1}\left(\frac{-2 - 1}{\sqrt{3}}\right) \right)$
$= \frac{1}{\sqrt{3}} \left( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(-\sqrt{3}) \right)$

6. Substitute exact trigonometric values and sum both parts to find $p$ and $q$.

Substitute the exact values $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ and $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$:
$= \frac{1}{\sqrt{3}} \left( \frac{\pi}{6} - \left(-\frac{\pi}{3}\right) \right) = \frac{1}{\sqrt{3}} \left( \frac{\pi}{6} + \frac{2\pi}{6} \right) = \frac{1}{\sqrt{3}} \left( \frac{3\pi}{6} \right) = \frac{\pi}{2\sqrt{3}} = \frac{\sqrt{3}}{6}\pi$

Adding both evaluated parts gives the total integral:
$\int_{-2}^{2} \frac{|x - 2|}{x^2 - 2x + 4} \,\mathrm{d}x = \frac{1}{2} \ln 3 + \frac{\sqrt{3}}{6}\pi$

Comparing this with $p \ln 3 + q \pi$, we obtain $p = \frac{1}{2}$ and $q = \frac{\sqrt{3}}{6}$.