By writing $x = A(8x - 8) + B$, where $A$ and $B$ are constants, find $\int_{0}^{1} \frac{x}{4x^2 - 8x + 5} \,\text{d}x$, giving your answer in the form $p \ln q + r \tan^{-1} s$, where $p$, $q$, $r$ and $s$ are values to be determined.

1. Determine the constants $A$ and $B$ by comparing coefficients.

$x = A(8x - 8) + B$
$x = 8Ax - 8A + B$
Comparing the coefficients of $x$:
$8A = 1 \implies A = \frac{1}{8}$
Comparing the constant terms:
$-8A + B = 0$
$-8\left(\frac{1}{8}\right) + B = 0 \implies B = 1$

2. Rewrite the integral by splitting it into two parts using the partial numerator.

Substitute $A$ and $B$ back into the integrand:
$\int_{0}^{1} \frac{x}{4x^2 - 8x + 5} \,\text{d}x = \int_{0}^{1} \frac{\frac{1}{8}(8x - 8) + 1}{4x^2 - 8x + 5} \,\text{d}x$
$= \frac{1}{8} \int_{0}^{1} \frac{8x - 8}{4x^2 - 8x + 5} \,\text{d}x + \int_{0}^{1} \frac{1}{4x^2 - 8x + 5} \,\text{d}x$

3. Evaluate the first integral using the standard form $\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)} \,\text{d}x$.

$\frac{1}{8} \int_{0}^{1} \frac{8x - 8}{4x^2 - 8x + 5} \,\text{d}x = \frac{1}{8} \left[ \ln |4x^2 - 8x + 5| \right]_0^1$
$= \frac{1}{8} \left( \ln(4(1)^2 - 8(1) + 5) - \ln(4(0)^2 - 8(0) + 5) \right)$
$= \frac{1}{8} (\ln 1 - \ln 5) = -\frac{1}{8} \ln 5$

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

$4x^2 - 8x + 5 = 4(x^2 - 2x) + 5$
$= 4(x^2 - 2x + 1) - 4 + 5$
$= 4(x - 1)^2 + 1$

5. Evaluate the second integral using the standard form $\int \frac{1}{x^2 + a^2} \,\text{d}x$ and combine.

$\int_{0}^{1} \frac{1}{4(x - 1)^2 + 1} \,\text{d}x = \frac{1}{4} \int_{0}^{1} \frac{1}{(x - 1)^2 + \left(\frac{1}{2}\right)^2} \,\text{d}x$
$= \frac{1}{4} \left[ \frac{1}{\frac{1}{2}} \tan^{-1} \left( \frac{x - 1}{\frac{1}{2}} \right) \right]_0^1 = \frac{1}{2} \left[ \tan^{-1}(2x - 2) \right]_0^1$
$= \frac{1}{2} (\tan^{-1}(0) - \tan^{-1}(-2))$
Since $\tan^{-1}(-x) = -\tan^{-1}(x)$, we have:
$= \frac{1}{2} (0 - (-\tan^{-1} 2)) = \frac{1}{2} \tan^{-1} 2$

Combining both parts:
$\int_{0}^{1} \frac{x}{4x^2 - 8x + 5} \,\text{d}x = -\frac{1}{8} \ln 5 + \frac{1}{2} \tan^{-1} 2$