Answer: (a) \(3(3x+1)\bigl(1+2\ln(3x+1)\bigr)\); (b) \(\frac16(3x+1)^2\ln(3x+1)-\frac34x^2-\frac12x+C\).

Set up the integral carefully. For an area between two curves, integrate the upper function minus the lower function over the correct interval.

(a) Use the product rule on

\(y=(3x+1)^2\ln(3x+1).\)

Then

\(\frac{\mathrm dy}{\mathrm dx} =6(3x+1)\ln(3x+1)+(3x+1)^2\cdot\frac{3}{3x+1}.\)

So

\(\frac{\mathrm dy}{\mathrm dx} =6(3x+1)\ln(3x+1)+3(3x+1).\)

Factorising,

\(\frac{\mathrm dy}{\mathrm dx} =3(3x+1)\bigl(1+2\ln(3x+1)\bigr).\)

(b) From part (a),

\(\frac{\mathrm dy}{\mathrm dx} =6(3x+1)\ln(3x+1)+3(3x+1).\)

Therefore

\((3x+1)\ln(3x+1) =\frac16\frac{\mathrm dy}{\mathrm dx}-\frac12(3x+1).\)

Integrate both sides:

\(\int(3x+1)\ln(3x+1)\,\mathrm dx =\frac16y-\frac12\int(3x+1)\,\mathrm dx.\)

Substitute \(y=(3x+1)^2\ln(3x+1)\):

Hence

\(\int(3x+1)\ln(3x+1)\,\mathrm dx =\frac16(3x+1)^2\ln(3x+1)-\frac34x^2-\frac12x+C.\)