= Solution
The required range is
$$
\boxed{x>-1.}
$$
Near $t=1$, the ratio $(t^x-1)/\ln t$ tends to $x$, so it causes no convergence problem. Near $t=0$, one useful identity is
$$
\frac{t^x-1}{\ln t}=\int_0^x t^s\,ds.
$$
For $x>-1$, this bounds its <absolute value> by $|x|t^{\min(x,0)}$, which is integrable on $(0,1)$. It also supplies the domination used in part (i).
For $x=-1$, the integrand is negative and behaves as $-1/[t|\ln t|]$ near zero; its <improper integral> diverges. For $x<-1$ its negative singularity is even stronger, and comparison with that same divergent <integral> again rules out convergence. Thus the original finite real-valued definition of $g$ has precisely the stated domain.
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