Show that $\frac{\text{d}}{\text{d}x}(2x \ln x) = 2 \ln x + 2$. A curve is such that the gradient of its tangent is $\ln x$ and it passes through the point $(e^2, 4)$. Using part (i), find the equation of the curve, leaving your answer in exact form.

1. Apply the product rule to differentiate $2x \ln x$

Let $u = 2x$ and $v = \ln x$.
$\frac{\text{d}u}{\text{d}x} = 2$
$\frac{\text{d}v}{\text{d}x} = \frac{1}{x}$
$\frac{\text{d}}{\text{d}x}(uv) = u\frac{\text{d}v}{\text{d}x} + v\frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(2x \ln x) = (2x)\left(\frac{1}{x}\right) + (\ln x)(2)$

2. Simplify the expression to obtain the required result

$= 2 + 2\ln x$
$= 2\ln x + 2$ (Shown)

1. Set up the integral for the equation of the curve

Given $\frac{\text{d}y}{\text{d}x} = \ln x$, we have:
$y = \int \ln x \, \text{d}x$

2. Use the result from part (i) to integrate $\ln x$

From part (i), $\frac{\text{d}}{\text{d}x}(2x \ln x) = 2 \ln x + 2$.
Integrating both sides with respect to $x$:
$2x \ln x = \int (2 \ln x + 2) \, \text{d}x$
$2x \ln x = 2 \int \ln x \, \text{d}x + \int 2 \, \text{d}x$
$2x \ln x = 2 \int \ln x \, \text{d}x + 2x + C'$
$2 \int \ln x \, \text{d}x = 2x \ln x - 2x - C'$
$\int \ln x \, \text{d}x = x \ln x - x + C$
So, $y = x \ln x - x + C$

3. Find the constant of integration using the given point

Substitute $x = e^2$ and $y = 4$ into the equation:
$4 = e^2 \ln(e^2) - e^2 + C$
$4 = e^2(2) - e^2 + C$
$4 = 2e^2 - e^2 + C$
$4 = e^2 + C$
$C = 4 - e^2$

4. State the final equation of the curve

$y = x \ln x - x + 4 - e^2$