答案是e - 2
∫(1->e) (lnx)² dx
= x(lnx)² |(1->e) - ∫(1->e) x d(lnx)² <= 分部积分法
= e - ∫(1->e) x * 2lnx * 1/x dx
= e - 2∫(1->e) lnx dx
= e - 2[xlnx |(1->e) - ∫(1->e) x d(lnx)] <= 分部积分法
= e - 2e + 2∫(1->e) x * 1/x dx
= -e + 2x |(1->e)
= -e + 2(e - 1)
= e - 2