Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 20 / c / iii

At $x=0$, the integrand vanishes for $0<t<1$, so $I(0)=0$ andSolving the first-order linear differential equation from part (i) on $(-1,\infty )$ gives $g=C(x+1)$. The initial condition fixes $C=1$, soEquivalently, the defining exponent is the logarithmic integral representation of the natural logarithm, $I(x)=\ln (x+1)$.