Let I_n = ∫_1^e x(lnx)ⁿ dx, where n ≥ 1.

Mark scheme

Marking points from the official mark scheme (May/June 2010). The mark or mark type is shown in brackets.

1. [2] (a) Apply integration by parts to I_n: ∫x(lnx)ⁿ dx = [(x²/2)(lnx)ⁿ]_1^e - ∫ (nx/2)(lnx)ⁿ⁻¹ dx
2. [1] (a) Evaluate and rearrange: I_n = e²/2 – (n/2)I_(n-1)
3. [1] (b) State inductive hypothesis H_k: I_k = A_k e² + B_k, where A_k and B_k are rational
4. [1] (b) Use recurrence relation for I_(k+1): I_(k+1) = e²/2 – ((k+1)/2)I_k
5. [1] (b) Substitute H_k into I_(k+1) and show A_(k+1) and B_(k+1) are rational: I_(k+1) = e²/2 – ((k+1)/2)(A_k e² + B_k) = (1/2 - (k+1)/2 A_k)e² - (k+1)/2 B_k
6. [2] (b) Base case H₁: I₁ = ∫_1^e x lnx dx = [x²/2 lnx]_1^e - ∫_1^e x²/2 (1/x) dx = e²/2 - [x²/4]_1^e = e²/2 - (e²/4 - 1/4) = e²/4 + 1/4. So A₁ = 1/4, B₁ = 1/4, which are rational. Thus H₁ is true.
7. [1] (b) Completion of induction argument