= Solution
{c}
There is no singularity at the lower endpoint one. Put $t=\log(1+x)$, so $dx=e^t\,dt$. The nonnegative integral becomes
$$
\int_{\log2}^\infty\frac{e^{(1-\alpha)t}}{t^2}\,dt.
$$
For $\alpha=1$, it converges as a reciprocal-square integral. For $\alpha>1$, exponential decay makes it converge. For $\alpha<1$, its integrand eventually grows without bound, so its integral diverges. Thus
$$
\boxed{\alpha\geq1.}
$$
This is the logarithmic borderline in a power-tail <integrability> test.
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