In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 14 - 2022 - Paper 1

Now, calculate the remaining integral:

$\int x^2 \, dx = \frac{x^3}{3}$

Substituting this back, we have: $\int \frac{x^2}{4} \ln{x} \, dx = \frac{x^3}{12} \ln{x} - \frac{1}{12} \cdot \frac{x^3}{3} + C$ Which simplifies to: $\int \frac{x^2}{4} \ln{x} \, dx = \frac{x^3}{12} \ln{x} - \frac{x^3}{36} + C$

Rearranging gives: $\int \frac{x^2}{4} \ln{x} \, dx = \frac{x^3}{12} \ln{x} -\frac{1}{36} x^3 + C$

To express it in the desired form: $ax^2 + b$ Let:

• $a = \frac{1}{12}$
• $b = -\frac{1}{36}$ thus, $\int_{e}^{2} \frac{x^2}{4} \ln{x} \, dx = ax^e + b$