Let $f(x)=\left(\frac{x}{2}\right)^x,\qquad g(x)=\left(\frac{2}{x}\right)^x.$ We need to evaluate $I=\int \left[f(x)+g(x)\right]\ln\left(\frac{ex}{2}\right)\,dx.$

Differentiate $f(x)=\left(\frac{x}{2}\right)^x$ using logarithmic differentiation: $\ln f=x\ln\left(\frac{x}{2}\right).$ Differentiating, $\frac{f'}{f}=\ln\left(\frac{x}{2}\right)+x\cdot \frac{1}{x}=\ln\left(\frac{x}{2}\right)+1.$ Hence, $f'(x)=\left(\frac{x}{2}\right)^x\left[\ln\left(\frac{x}{2}\right)+1\right].$ But $\ln\left(\frac{x}{2}\right)+1=\ln\left(e\cdot \frac{x}{2}\right)=\ln\left(\frac{ex}{2}\right).$ So, $f'(x)=\left(\frac{x}{2}\right)^x\ln\left(\frac{ex}{2}\right).$

Now differentiate $g(x)=\left(\frac{2}{x}\right)^x$: $\ln g=x\ln\left(\frac{2}{x}\right).$ Differentiating, $\frac{g'}{g}=\ln\left(\frac{2}{x}\right)+x\left(-\frac{1}{x}\right)=\ln\left(\frac{2}{x}\right)-1.$ Thus, $g'(x)=\left(\frac{2}{x}\right)^x\left[\ln\left(\frac{2}{x}\right)-1\right].$ Now, $\ln\left(\frac{2}{x}\right)-1=\ln\left(\frac{2}{x}\right)-\ln e=\ln\left(\frac{2}{ex}\right)=-\ln\left(\frac{ex}{2}\right).$ Therefore, $g'(x)=-\left(\frac{2}{x}\right)^x\ln\left(\frac{ex}{2}\right).$

Hence, $-g'(x)=\left(\frac{2}{x}\right)^x\ln\left(\frac{ex}{2}\right).$ So the integrand becomes $\left(\frac{x}{2}\right)^x\ln\left(\frac{ex}{2}\right)+\left(\frac{2}{x}\right)^x\ln\left(\frac{ex}{2}\right)=f'(x)-g'(x).$ Thus, $I=\int \left[f'(x)-g'(x)\right]dx=f(x)-g(x)+C.$

Substitute back: $I=\left(\frac{x}{2}\right)^x-\left(\frac{2}{x}\right)^x+C.$

Comparing with the options, this is Option B.