= Solution
For $x>1$ the geometric expansion gives
$$
\frac{\log x}{x^p(x-1)}=\sum_{k=0}^\infty x^{-(p+k+1)}\log x.
$$
Every summand is nonnegative, so <monotone convergence> permits integration term by term without a prior integrability assumption. The preceding part applies with $\alpha=p+k+1>1$, yielding
$$
\boxed{\int_1^\infty\frac{\log x}{x^p(x-1)}\,dx=\sum_{k=0}^\infty\frac1{(p+k)^2}.}
$$
The series is finite for every fixed $p>0$, by comparison of its tail with $\sum k^{-2}$. This also proves integrability of the original function.
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