The integral $\\int_{2}^{3} x^{3} \\ln(2x) \, dx$ can be written in the form $p \\ln 2 + q$, where $p$ and $q$ are rational numbers - AQA - A-Level Maths Pure - Question 4 - 2019 - Paper 3

Differentiating and integrating gives us:

• $du = \frac{2}{x} \, dx$
• $v = \frac{x^{4}}{4}$

Substituting into the integration by parts formula: $\int x^{3} \\ln(2x) \, dx = \left( \ln(2x) \cdot \frac{x^{4}}{4} \right) - \int \frac{x^{4}}{4} \cdot \frac{2}{x} \, dx$ This simplifies to: $\int x^{3} \\ln(2x) \, dx = \left( \ln(2x) \cdot \frac{x^{4}}{4} \right) - \int \frac{x^{3}}{2} \, dx$

Now, we compute the integral: $\int \frac{x^{3}}{2} \, dx = \frac{x^{4}}{8}$ Thus, we have: $\int x^{3} \\ln(2x) \, dx = \frac{x^{4}}{4} \ln(2x) - \frac{x^{4}}{16} + C$ This can be rewritten as: $\frac{x^{4}}{4} \ln(2) + \frac{x^{4}}{4} \ln(x) - \frac{x^{4}}{16} + C$

Substituting the limits from $2$ to $3$, we find:

1. When $x = 3$: $\frac{3^{4}}{4} \ln(6) - \frac{3^{4}}{16}$
2. When $x = 2$: $\frac{2^{4}}{4} \ln(4) - \frac{2^{4}}{16}$ Evaluating these, we find:

• $p = \frac{3^{4}}{4} - \frac{2^{4}}{4} = \frac{81}{4} - 4 = \frac{81 - 16}{4} = \frac{65}{4}$
• $q = \left(-\frac{81}{16} + 4\right) - \left(-\frac{16}{16} + 1\right)$ Calculating gives us the values of $p$ and $q$.