Key Concepts and Formulas

• Derivative of $x^x$ type functions: If $y = u(x)^{v(x)}$, then $\frac{dy}{dx} = u(x)^{v(x)} \left[ v'(x) \ln(u(x)) + \frac{v(x)u'(x)}{u(x)} \right]$
• Indefinite Integral: $\int f'(x) \, dx = f(x) + C$, where C is the constant of integration.
• Logarithm Properties: $\ln(ab) = \ln(a) + \ln(b)$, $\ln(a/b) = \ln(a) - \ln(b)$, $\ln(e) = 1$

Step-by-Step Solution

Step 1: Simplify the logarithmic term

The goal is to simplify the integrand to make it easier to work with. We start by simplifying the logarithmic term using logarithm properties. $\ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right)$

Step 2: Rewrite the integral

Now we substitute the simplified logarithmic term back into the integral. $I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \left(1 + \ln \left(\frac{x}{2}\right)\right) d x$

Step 3: Split the integral

We split the integral into two parts to make it easier to manage. $I = \int \left(\frac{x}{2}\right)^x \left(1 + \ln \left(\frac{x}{2}\right)\right) d x + \int \left(\frac{2}{x}\right)^x \left(1 + \ln \left(\frac{x}{2}\right)\right) d x$

Step 4: Evaluate the first integral

Let $A = \left(\frac{x}{2}\right)^x$. We want to find $\frac{dA}{dx}$. Using the formula for the derivative of $u(x)^{v(x)}$, where $u(x) = \frac{x}{2}$ and $v(x) = x$: $\frac{dA}{dx} = \left(\frac{x}{2}\right)^x \left(1 \cdot \ln\left(\frac{x}{2}\right) + x \cdot \frac{1}{x/2} \cdot \frac{1}{2}\right) = \left(\frac{x}{2}\right)^x \left(\ln\left(\frac{x}{2}\right) + 1\right)$ So, $\int \left(\frac{x}{2}\right)^x \left(1 + \ln \left(\frac{x}{2}\right)\right) d x = \int \frac{dA}{dx} dx = A + C_1 = \left(\frac{x}{2}\right)^x + C_1$

Step 5: Evaluate the second integral

Let $B = \left(\frac{2}{x}\right)^x$. We want to find $\frac{dB}{dx}$. Using the formula for the derivative of $u(x)^{v(x)}$, where $u(x) = \frac{2}{x}$ and $v(x) = x$: $\frac{dB}{dx} = \left(\frac{2}{x}\right)^x \left(1 \cdot \ln\left(\frac{2}{x}\right) + x \cdot \frac{-2}{x^2} \cdot \frac{x}{2} \right) = \left(\frac{2}{x}\right)^x \left(\ln\left(\frac{2}{x}\right) - 1\right)$ We have $1 + \ln\left(\frac{x}{2}\right) = 1 + \ln(x) - \ln(2)$. Also, $\ln\left(\frac{2}{x}\right) = \ln(2) - \ln(x)$. Therefore, $1 + \ln\left(\frac{x}{2}\right) = -( \ln\left(\frac{2}{x}\right) -1 )$. So we have $\int \left(\frac{2}{x}\right)^x \left(1 + \ln \left(\frac{x}{2}\right)\right) d x = \int \left(\frac{2}{x}\right)^x \left(- \ln \left(\frac{2}{x}\right) +1 \right)(-1)dx = - \int \frac{dB}{dx} dx = -B + C_2 = - \left(\frac{2}{x}\right)^x + C_2$

Step 6: Combine the results

Add the two integrals: $I = \left(\frac{x}{2}\right)^x - \left(\frac{2}{x}\right)^x + C$

Common Mistakes & Tips

• Carefully apply the chain rule when differentiating $u(x)^{v(x)}$.
• Pay attention to the signs when dealing with logarithms and derivatives.
• Remember to add the constant of integration, $C$, to the final result.

Summary

We simplified the integral by using logarithm properties, splitting the integral into two parts, and then recognizing that each part could be solved by finding the derivative of a function. By careful application of the chain rule and logarithm properties, we found that the integral evaluates to $\left(\frac{x}{2}\right)^x - \left(\frac{2}{x}\right)^x + C$.

Final Answer

The final answer is \boxed{\left(\frac{x}{2}\right)^{x}-\left(\frac{2}{x}\right)^{x}+C}, which corresponds to option (B).