Express $\frac{8x^2 - x}{4x^2 - 1}$ in partial fractions. Use your results in part (i) to find $\int \frac{8x^2 - x}{4x^2 - 1} \, dx$.

1. Perform long division since the degree of the numerator is equal to the degree of the denominator.

$\frac{8x^2 - x}{4x^2 - 1} = \frac{2(4x^2 - 1) - x + 2}{4x^2 - 1} = 2 + \frac{2 - x}{4x^2 - 1}$

2. Factorise the denominator and set up the partial fraction decomposition.

$\frac{2 - x}{4x^2 - 1} = \frac{2 - x}{(2x - 1)(2x + 1)}$
Let $\frac{2 - x}{(2x - 1)(2x + 1)} = \frac{A}{2x - 1} + \frac{B}{2x + 1}$

3. Solve for the unknown constants $A$ and $B$.

$2 - x = A(2x + 1) + B(2x - 1)$
Substitute $x = \frac{1}{2}$:
$2 - \frac{1}{2} = A(2(\frac{1}{2}) + 1) \implies \frac{3}{2} = 2A \implies A = \frac{3}{4}$
Substitute $x = -\frac{1}{2}$:
$2 - (-\frac{1}{2}) = B(2(-\frac{1}{2}) - 1) \implies \frac{5}{2} = -2B \implies B = -\frac{5}{4}$

4. Write the final expression in partial fractions.

$\frac{8x^2 - x}{4x^2 - 1} = 2 + \frac{3}{4(2x - 1)} - \frac{5}{4(2x + 1)}$

1. Substitute the partial fraction decomposition from part (i)

From part (i), we have:
$\frac{8x^2 - x}{4x^2 - 1} = 2 + \frac{3}{4(2x - 1)} - \frac{5}{4(2x + 1)}$

$\int \frac{8x^2 - x}{4x^2 - 1} \, dx = \int \left( 2 + \frac{3}{4(2x - 1)} - \frac{5}{4(2x + 1)} \right) \, dx$

2. Integrate each term using the standard integral for linear denominators

$= 2x + \frac{3}{4} \left( \frac{\ln(2x - 1)}{2} \right) - \frac{5}{4} \left( \frac{\ln(2x + 1)}{2} \right) + c$
$= 2x + \frac{3}{8} \ln(2x - 1) - \frac{5}{8} \ln(2x + 1) + c$