Use integration by parts: Let \( u = (\ln x)^2 \), \( dv = x \, dx \). M1

Then: \( du = 2 \ln x \cdot \frac{1}{x} \, dx = \frac{2 \ln x}{x} \, dx \), \( v = \frac{x^2}{2} \). A1

Apply integration by parts: \(\int u \, dv = uv – \int v \, du\).

\(\int x (\ln x)^2 \, dx = \frac{x^2}{2} (\ln x)^2 – \int \frac{x^2}{2} \cdot \frac{2 \ln x}{x} \, dx = \frac{x^2}{2} (\ln x)^2 – \int x \ln x \, dx\). A1

For \(\int x \ln x \, dx\), use integration by parts again: Let \( u = \ln x \), \( dv = x \, dx \). M1

Then: \( du = \frac{1}{x} \, dx \), \( v = \frac{x^2}{2} \). A1

\(\int x \ln x \, dx = \frac{x^2}{2} \ln x – \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x – \frac{1}{2} \int x \, dx = \frac{x^2}{2} \ln x – \frac{x^2}{4} + C_1\). A1

Substitute back: \(\int x (\ln x)^2 \, dx = \frac{x^2}{2} (\ln x)^2 – \left( \frac{x^2}{2} \ln x – \frac{x^2}{4} \right) = \frac{x^2}{2} (\ln x)^2 – \frac{x^2}{2} \ln x + \frac{x^2}{4}\).

Combine: \(\frac{x^2}{2} \left( (\ln x)^2 – \ln x + \frac{1}{2} \right) + C\). A1

[6 marks]

Evaluate \(\int_{1}^{4} x (\ln x)^2 \, dx = \left[ \frac{x^2}{2} \left( (\ln x)^2 – \ln x + \frac{1}{2} \right) \right]_{1}^{4}\). M1

Upper limit \( x = 4 \): \(\ln 4 = 2 \ln 2\), \((\ln 4)^2 = (2 \ln 2)^2 = 4 (\ln 2)^2\).

\(\frac{4^2}{2} \left( 4 (\ln 2)^2 – 2 \ln 2 + \frac{1}{2} \right) = 8 \left( 4 (\ln 2)^2 – 2 \ln 2 + \frac{1}{2} \right) = 32 (\ln 2)^2 – 16 \ln 2 + 4\). A1

Lower limit \( x = 1 \): \(\ln 1 = 0\), \(\frac{1^2}{2} \left( 0 – 0 + \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\). A1

Combine: \(32 (\ln 2)^2 – 16 \ln 2 + 4 – \frac{1}{4} = 32 (\ln 2)^2 – 16 \ln 2 + \frac{16}{4} – \frac{1}{4} = 32 (\ln 2)^2 – 16 \ln 2 + \frac{15}{4}\). A1

[3 marks]

Total [9 marks]