Given that $y = (2x - 1)^3 \ln \sqrt{2x - 1}$. Show that $\frac{\text{d}y}{\text{d}x} = 6(2x - 1)^2 \ln \sqrt{2x - 1} + (2x - 1)^2$. Hence evaluate $\int_{1}^{5} (2x - 1)^2 \ln \sqrt{2x - 1} \text{ d}x$.

1. Simplify the logarithmic term using laws of logarithms before differentiating

$y = (2x - 1)^3 \ln (2x - 1)^{\frac{1}{2}}$
$y = \frac{1}{2}(2x - 1)^3 \ln (2x - 1)$

2. Apply the product rule to differentiate the expression

Let $u = \frac{1}{2}(2x - 1)^3$ and $v = \ln (2x - 1)$.
$\frac{\text{d}u}{\text{d}x} = \frac{1}{2} \cdot 3(2x - 1)^2 \cdot 2 = 3(2x - 1)^2$
$\frac{\text{d}v}{\text{d}x} = \frac{1}{2x - 1} \cdot 2 = \frac{2}{2x - 1}$
$\frac{\text{d}y}{\text{d}x} = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}y}{\text{d}x} = \frac{1}{2}(2x - 1)^3 \left( \frac{2}{2x - 1} \right) + \ln (2x - 1) \left( 3(2x - 1)^2 \right)$

3. Simplify the derivative to match the required form

$\frac{\text{d}y}{\text{d}x} = (2x - 1)^2 + 3(2x - 1)^2 \ln (2x - 1)$
Since $\ln (2x - 1) = 2 \ln (2x - 1)^{\frac{1}{2}} = 2 \ln \sqrt{2x - 1}$:
$\frac{\text{d}y}{\text{d}x} = (2x - 1)^2 + 3(2x - 1)^2 (2 \ln \sqrt{2x - 1})$
$\frac{\text{d}y}{\text{d}x} = 6(2x - 1)^2 \ln \sqrt{2x - 1} + (2x - 1)^2$ (Shown)