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