int {{e^x} left[ frac{1 + x log x}{x} right] | vedclass

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(D) We know that the integral of the form $\int {{e^x} [f(x) + f'(x)] \, dx} = {e^x} f(x) + c$. Given integral is $\int {{e^x} \left[ \frac{1 + x \log

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${e^x} \log x + c$

Vedclass · Answer

(D) We know that the integral of the form $\int {{e^x} [f(x) + f'(x)] \, dx} = {e^x} f(x) + c$. Given integral is $\int {{e^x} \left[ \frac{1 + x \log x}{x} \right] \, dx}$. This can be rewritten as $\int {{e^x} \left( \frac{1}{x} + \log x \right) \, dx}$. Here,let $f(x) = \log x$,then $f'(x) = \frac{1}{x}$. Substituting this into the standard formula,we get $\int {{e^x} [\log x + \frac{1}{x}] \, dx} = {e^x} \log x + c$.

$\int {{e^x} \left[ \frac{1 + x \log x}{x} \right] \, dx} = $

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