= Solution
{c}
For each nonnegative <integer> $k$, the <change of variables> $x=e^{-t}$ gives
$$
\int_0^1|(x\log x)^k|\,dx=\int_0^\infty t^ke^{-(k+1)t}\,dt
=\frac{\Gamma(k+1)}{(k+1)^{k+1}}<\infty.
$$
The <gamma function> identity $\Gamma(k+1)=k!$ follows by repeated <integration by parts>. Since $\log x<0$ on $(0,1)$, the sign of the original integrand is $(-1)^k$, so
$$
\boxed{\int_0^1(x\log x)^k\,dx=\frac{(-1)^kk!}{(k+1)^{k+1}}.}
$$
For $k=0$, the integral is one, consistent with the formula.
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