= Solution
For $1<x<e$, the <natural logarithm> satisfies $0<\ln x<1$. Squaring therefore decreases its value, while $\ln(x^2)=2\ln x$. Pointwise comparison gives $(\ln x)^2<\ln x<\ln(x^2)$, so
$$
\boxed{I_3<I_1<I_2.}
$$
As a check, <integration by parts> gives $I_1=1$, $I_2=2$ and $I_3=e-2$, consistent with this order.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1-logarithm-comparison.png]
{title=Pointwise ordering of the three logarithmic integrands}
{description=Original plot on $[1,e]$. The natural logarithm rises from zero to one, so its square lies below it, and twice the logarithm lies above it. Integrating preserves the strict order.}
Back to article page
