Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 4 / c

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 $(\ln x{)}^2<\ln x<\ln (x^2)$, soAs a check, integration by parts gives $I_1=1$, $I_2=2$ and $I_3=e-2$, consistent with this order.