计算 $\int (x+2024)\ln x\,dx=\int x\ln x\,dx+2024\int \ln x\,dx$。先求 $\int x\ln x\,dx$：令 $u=\ln x$，$dv=x\,dx$，则 $du=\dfrac{1}{x}dx$，$v=\dfrac{1}{2}x^2$，得 $\int x\ln x\,dx=\dfrac{1}{2}x^2\ln x-\int \dfrac{1}{2}x^2\cdot\dfrac{1}{x}dx=\dfrac{1}{2}x^2\ln x-\dfrac{1}{2}\int x\,dx=\dfrac{1}{2}x^2\ln x-\dfrac{x^2}{4}$。再求 $\int \ln x\,dx=x\ln x-x$。故原积分为 $\dfrac{1}{2}x^2\ln x-\dfrac{x^2}{4}+2024(x\ln x-x)+C=\dfrac{1}{2}x^2\ln x-\dfrac{x^2}{4}+2024x\ln x-2024x+C$。