int {{e^x} left( {frac{1}{x} - frac{1}{{{x^2} | vedclass

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Question

(C) We use the standard integration formula: $\int {{e^x} \{ f(x) + f'(x) \} \,dx} = {e^x}f(x) + c$. Here,let $f(x) = \frac{1}{x}$. Then,the derivativ

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We use the standard integration formula: $\int {{e^x} \{ f(x) + f'(x) \} \,dx} = {e^x}f(x) + c$. Here,let $f(x) = \frac{1}{x}$. Then,the derivative $f'(x) = -\frac{1}{{{x^2}}}$. Substituting these into the formula,we get: $\int {{e^x} \left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} \,dx = {e^x} \left( \frac{1}{x} \right) + c = \frac{{{e^x}}}{x} + c$. Thus,the correct option is $C$.

$\int {{e^x} \left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} \,dx = $

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