解答
(1)
\begin{eqnarray}
(与式)&& = 2 \int_1^4 x^{\frac{1}{2}} \log x dx \\
&& = 2 \int_1^4 \frac{2}{3} \left( x^{\frac{3}{2}} \right)^{\prime} \log x dx \\
&& = \frac{4}{3} \left[ x^{\frac{3}{2}} \log x \right]_1^4 – \frac{4}{3} \int_1^4 x^{\frac{3}{2}} \cdot \frac{1}{x} dx \\
&& = \frac{4}{3} 2^3 \log 4 – \frac{4}{3} \int_1^4 x^{\frac{1}{2}} dx \\
&& = \frac{2^6}{3} \log 2 – \frac{4}{3} \frac{2}{3} [ x^{\frac{3}{2}} ]_1^4 \\
&& = \frac{64}{3} \log 2 – \frac{8}{9} (8-1) \\
&& = \frac{64}{3} \log 2 – \frac{56}{9} \cdots (答)
\end{eqnarray}
(2)
\begin{eqnarray}
&& x^{2023}-1 \\
&& \hspace{10pt} = (x-1)(x^{2022}+x^{2021}+x^{2020}+ \cdots +1) \\
&& \hspace{10pt} = (x-1) \{ x^{2018}(x^4+x^3+\cdots+1) + x^{2013}(x^4+x^3+\cdots+1) + \cdots \\
&& \hspace{20pt} +x^3(x^4+x^3+\cdots+1)+x^2+x+1 \} \\
&& \hspace{10pt} = (x-1)(x^{2018}+x^{2013}+\cdots+x^3)(x^4+x^3+\cdots+1) \\
&& \hspace{20pt} + (x-1)(x^2+x+1) \\
\end{eqnarray}
よって求める余りは
\begin{equation}
(x-1)(x^2+x+1) = x^3-1 \cdots (答)
\end{equation}