Given that $y = x^2 \sqrt{2x-1}$, show that $\frac{\text{d}y}{\text{d}x} = \frac{x(5x-2)}{\sqrt{2x-1}}$. Determine whether $y = x^2 \sqrt{2x-1}$ is always an increasing function. Hence evaluate $\int_{1}^{5} \frac{5x^2 - 2x + 1}{\sqrt{2x-1}} \text{d}x$.

1. Differentiate the function using the product rule and chain rule.

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

$\frac{\text{d}y}{\text{d}x} = 2x(2x-1)^{\frac{1}{2}} + x^2 \left[ \frac{1}{2}(2x-1)^{-\frac{1}{2}}(2) \right]$
$\frac{\text{d}y}{\text{d}x} = 2x(2x-1)^{\frac{1}{2}} + x^2(2x-1)^{-\frac{1}{2}}$

2. Factorise and simplify the expression to show the required result.

$\frac{\text{d}y}{\text{d}x} = x(2x-1)^{-\frac{1}{2}} \left[ 2(2x-1) + x \right]$
$\frac{\text{d}y}{\text{d}x} = \frac{x(4x - 2 + x)}{\sqrt{2x-1}}$
$\frac{\text{d}y}{\text{d}x} = \frac{x(5x-2)}{\sqrt{2x-1}}$ (Shown)

3. Determine the domain of the function and analyse the sign of the derivative.

For $y$ to be defined, $2x - 1 \geqslant 0 \implies x \geqslant \frac{1}{2}$.
For $x > \frac{1}{2}$:
$x > 0$
$5x - 2 > 5(\frac{1}{2}) - 2 = 0.5 > 0$
$\sqrt{2x-1} > 0$
Since all terms in $\frac{x(5x-2)}{\sqrt{2x-1}}$ are positive for $x > \frac{1}{2}$, $\frac{\text{d}y}{\text{d}x} > 0$.
Therefore, $y$ is always an increasing function on its domain.

1. Relate the integrand to the derivative from part (a)

From (a), we know that $\frac{\text{d}}{\text{d}x} (x^2 \sqrt{2x-1}) = \frac{5x^2 - 2x}{\sqrt{2x-1}}$.
We want to evaluate $\int_{1}^{5} \frac{5x^2 - 2x + 1}{\sqrt{2x-1}} \text{d}x$.
Notice that $\frac{5x^2 - 2x + 1}{\sqrt{2x-1}} = \frac{5x^2 - 2x}{\sqrt{2x-1}} + \frac{1}{\sqrt{2x-1}}$.

2. Express the integral as the sum of two integrals

$\int_{1}^{5} \frac{5x^2 - 2x + 1}{\sqrt{2x-1}} \text{d}x = \int_{1}^{5} \frac{5x^2 - 2x}{\sqrt{2x-1}} \text{d}x + \int_{1}^{5} (2x-1)^{-\frac{1}{2}} \text{d}x$

3. Integrate both terms

$\int_{1}^{5} \frac{5x^2 - 2x}{\sqrt{2x-1}} \text{d}x = \left[ x^2 \sqrt{2x-1} \right]_{1}^{5}$
$\int_{1}^{5} (2x-1)^{-\frac{1}{2}} \text{d}x = \left[ \frac{(2x-1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 2} \right]_{1}^{5} = \left[ \sqrt{2x-1} \right]_{1}^{5}$

4. Evaluate the definite integral

$\left[ x^2 \sqrt{2x-1} + \sqrt{2x-1} \right]_{1}^{5} = \left[ (x^2 + 1)\sqrt{2x-1} \right]_{1}^{5}$
Substitute $x = 5$:
$(5^2 + 1)\sqrt{2(5)-1} = 26 \sqrt{9} = 26(3) = 78$
Substitute $x = 1$:
$(1^2 + 1)\sqrt{2(1)-1} = 2 \sqrt{1} = 2$
Evaluate:
$78 - 2 = 76$