Let f(x) = frac{1}{x} ln left( frac{x}{e^x} r | vedclass

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(A) We are given $f(x) = \frac{1}{x} \ln \left( \frac{x}{e^x} \right)$.To find the primitive (indefinite integral),we calculate $\int f(x) \, dx = \in

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$\frac{1}{2} \ln^2 x - x + C$

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(A) We are given $f(x) = \frac{1}{x} \ln \left( \frac{x}{e^x} \right)$.To find the primitive (indefinite integral),we calculate $\int f(x) \, dx = \int \frac{1}{x} \ln \left( \frac{x}{e^x} \right) \, dx$.Using the logarithmic property $\ln \left( \frac{a}{b} \right) = \ln a - \ln b$,we get:$\int \frac{1}{x} (\ln x - \ln e^x) \, dx = \int \frac{\ln x - x}{x} \, dx$.This simplifies to $\int \left( \frac{\ln x}{x} - 1 \right) \, dx$.Splitting the integral: $\int \frac{\ln x}{x} \, dx - \int 1 \, dx$.For the first part,let $u = \ln x$,then $du = \frac{1}{x} \, dx$. The integral becomes $\int u \, du = \frac{u^2}{2} = \frac{(\ln x)^2}{2}$.For the second part,$\int 1 \, dx = x$.Combining these,we get $\frac{1}{2} \ln^2 x - x + C$.

Let $f(x) = \frac{1}{x} \ln \left( \frac{x}{e^x} \right)$,then its primitive with respect to $x$ is:

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