[Cambridge/9709/2007] Use integration by parts to show that
$\int_{2}^{4} \ln x \, dx = 6 \ln 2 - 2.$

1. Identify the parts $u$ and $dv$ for integration by parts.

Let $u = \ln x$ and $\frac{dv}{dx} = 1$.
Then $\frac{du}{dx} = \frac{1}{x}$ and $v = x$.

2. Apply the integration by parts formula to the definite integral.

Using the integration by parts formula $\int_{a}^{b} u \, dv = [uv]_{a}^{b} - \int_{a}^{b} v \, du$:
$\int_{2}^{4} \ln x \, dx = [x \ln x]_{2}^{4} - \int_{2}^{4} x \left(\frac{1}{x}\right) \, dx$
$= [x \ln x]_{2}^{4} - \int_{2}^{4} 1 \, dx$

3. Evaluate the definite integrals using the limits.

$= (4 \ln 4 - 2 \ln 2) - [x]_{2}^{4}$
$= 4 \ln 4 - 2 \ln 2 - (4 - 2)$
$= 4 \ln 4 - 2 \ln 2 - 2$

4. Simplify the logarithmic terms to show the final result.

Use the power law of logarithms, $\ln(a^n) = n \ln a$, to simplify $\ln 4$:
$\ln 4 = \ln(2^2) = 2 \ln 2$
Substitute this back into the expression:
$= 4(2 \ln 2) - 2 \ln 2 - 2$
$= 8 \ln 2 - 2 \ln 2 - 2$
$= 6 \ln 2 - 2 \text{ (shown)}$