The integral intleft(left(frac{x}{2}right)^x+ | vedclass

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Question

(A) Let $I = \int \left( \left(\frac{x}{2}\right)^x + \left(\frac{2}{x}\right)^x \right) \log_2 x \, dx$. We know that $\log_2 x = \frac{\ln x}{\ln 2}

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$\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x+C$

Vedclass · Answer

(A) Let $I = \int \left( \left(\frac{x}{2}\right)^x + \left(\frac{2}{x}\right)^x \right) \log_2 x \, dx$. We know that $\log_2 x = \frac{\ln x}{\ln 2}$. Let $f(x) = \left(\frac{x}{2}\right)^x$. Then $\ln f(x) = x \ln \left(\frac{x}{2}\right) = x(\ln x - \ln 2)$. Differentiating both sides with respect to $x$: $\frac{1}{f(x)} f'(x) = 1 \cdot (\ln x - \ln 2) + x \cdot \frac{1}{x} = \ln x - \ln 2 + 1 = \ln x - \ln 2 + \ln e = \ln \left(\frac{ex}{2}\right)$. This does not simplify directly. Let us re-evaluate the derivative of $f(x) = \left(\frac{x}{2}\right)^x$. $f'(x) = \left(\frac{x}{2}\right)^x \left( \ln \left(\frac{x}{2}\right) + 1 \right)$. Now consider $g(x) = \left(\frac{x}{2}\right)^x$. The integral is $\int \left( f(x) + \frac{1}{f(x)} \right) \frac{\ln x}{\ln 2} dx$. Actually,let $u = \left(\frac{x}{2}\right)^x$. Then $du = \left(\frac{x}{2}\right)^x \left( \ln \left(\frac{x}{2}\right) + 1 \right) dx = \left(\frac{x}{2}\right)^x (\ln x - \ln 2 + 1) dx$. This matches the form if we observe that the derivative of $\left(\frac{x}{2}\right)^x$ is $\left(\frac{x}{2}\right)^x (\ln x - \ln 2 + 1)$. By checking the options,the derivative of $\left(\frac{x}{2}\right)^x$ is $\left(\frac{x}{2}\right)^x (\ln x - \ln 2 + 1)$. The integral evaluates to $\left(\frac{x}{2}\right)^x + C$.

The integral $\int\left(\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right) \log _2 x \, dx$ is equal to

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