Zadanie 8a, Paper 1, listopad 2023, TZ1, Matematyka AA IB HL

Let I be the integral. Integration by parts with u=(lnx)²,dv=x²dx gives $I=x^3(\ln x)^2/3-(2/3)\int x^2\ln x\,dx$. A second application gives $\int x^2\ln x\,dx=x^3\ln x/3-x^3/9$. Therefore I = x³(ln x)²/3 − 2x³ln x/9 + 2x³/27 + C.

M1:attempt integration by parts. M1:suitable choice separating the logarithmic factor and power. A1:correct first reduced integral. M1:apply parts again to the remaining logarithmic integral. A1:correct second antiderivative. A1:complete primitive. Equivalent valid two-step parts choices earn corresponding marks. The additive constant may be omitted without penalty, as in the source rubric. If J=∫x²ln x dx=x³ln x/3−x³/9 is known, use u=ln x,dv=x²ln x dx: M1 for parts, M1 for this choice, A1 for I=ln x·J−∫J/x dx, A1 for the reduced integrand, A1 for its integration, A1 for the complete result. Other valid rearrangements that solve a recurrence for I are equally creditworthy.