int e^x left( log x + frac{1}{x} - frac{1}{x} | vedclass

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Question

(C) We know the standard integral formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Here,let $f(x) = \log x$.
Then,$f'(x) = \frac{1}{x}$

Vedclass · Accepted Answer

$e^x \log x + C$

Vedclass · Answer

(C) We know the standard integral formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Here,let $f(x) = \log x$.
Then,$f'(x) = \frac{1}{x}$.
Substituting these into the formula:
$\int e^x \left( \log x + \frac{1}{x} \right) dx = e^x \log x + C$.
However,looking at the provided options,the question likely intended to be $\int e^x \left( \log x + \frac{1}{x} \right) dx$. If the question is $\int e^x \left( \log x + \frac{1}{x} \right) dx$,the result is $e^x \log x + C$. If the question is $\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx$,the result is $e^x \frac{1}{x} + C$. Given the structure of the options,there is a mismatch. Assuming the standard form $\int e^x (f(x) + f'(x)) dx$,the closest match for the form $e^x(\log x + \dots)$ is option $C$ if the derivative was different.

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