Hence show that $\int x\ln x\,dx = \tfrac{1}{2}x^2\ln x - \tfrac{1}{4}x^2 + c$, where c is a constant. The curve $y = f(x)$ passes through the point $\left(1,\tfrac{19}{4}\right)$ and is such that $f'(x)=\tfrac{1}{x} + x\ln x$. Find $f(x)$. State the range of values of $x$ for which $f(x)$ is defined.

1. State the derivative of $x^2\ln x$ from part (i)

By the product rule,
$\frac{d}{dx}\left(x^2\ln x\right) = 2x\ln x + x^2\cdot\frac{1}{x} = 2x\ln x + x.$

2. Integrate both sides (reverse of differentiation)

Since integration reverses differentiation,
$\int \left(2x\ln x + x\right)dx = x^2\ln x + C.$
So
$2\int x\ln x\,dx + \int x\,dx = x^2\ln x + C.$

3. Rearrange to make $\int x\ln x\,dx$ the subject

$2\int x\ln x\,dx = x^2\ln x - \frac{1}{2}x^2 + C.$
Dividing by $2$:
$\int x\ln x\,dx = \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + c,$
where $c=\tfrac{C}{2}$ is a constant. (shown)

1. Integrate $f'(x)$ to find the general form of $f(x)$

$f(x) = \int \left(\frac{1}{x} + x\ln x\right)\,dx$
Using the standard integral for $\frac{1}{x}$ and the result from part (ii):
$f(x) = \ln x + \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + C$

2. Substitute the given point to find the constant of integration $C$

Substitute $x = 1$ and $f(1) = \frac{19}{4}$:
$\frac{19}{4} = \ln(1) + \frac{1}{2}(1)^2\ln(1) - \frac{1}{4}(1)^2 + C$

3. Solve for $C$ and state $f(x)$

Since $\ln(1) = 0$:
$\frac{19}{4} = 0 + 0 - \frac{1}{4} + C$
$C = \frac{19}{4} + \frac{1}{4} = \frac{20}{4} = 5$
$\therefore f(x) = \ln x + \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + 5$

1. Identify the domain restrictions for the function

The function $f(x)$ and its derivative $f'(x) = \frac{1}{x} + x\ln x$ contain the natural logarithm term $\ln x$, which is only defined for strictly positive values of $x$. The term $\frac{1}{x}$ also requires $x \neq 0$. Therefore, $x > 0$.