= Solution
Write $I(x)=\ln g(x)$. For $x>-1$, <differentiation under the integral sign> gives
$$
I'(x)=\int_0^1\frac{\partial}{\partial x}\left(\frac{t^x-1}{\ln t}\right)dt=\int_0^1t^x\,dt=\frac1{x+1}.
$$
Indeed, on any compact parameter <interval> $c\leq x\leq d$ with $c>-1$, the differentiated integrand is bounded by the integrable <function> $t^c$. The <dominated convergence theorem> therefore justifies the operation despite the improper lower endpoint. The apparent singularity at $t=1$ is removable.
Since $g=e^I$, the <chain rule> gives the <first-order linear differential equation>
$$
\boxed{g'(x)=\frac{g(x)}{x+1},\qquad x>-1.}
$$
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