int_1^2 {{e^x}left( {frac{1}{x} - frac{1}{{{x | vedclass

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Question

(C) We use the standard integral formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$. Here,let $f(x) = \frac{1}{x}$. Then,$f'(x) = -\frac{1}{x^2}$.

Vedclass · Accepted Answer

$\frac{{{e^2}}}{2} - e$

Vedclass · Answer

(C) We use the standard integral formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$. Here,let $f(x) = \frac{1}{x}$. Then,$f'(x) = -\frac{1}{x^2}$. Thus,the integral becomes $\int_1^2 e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) \, dx = \left[ e^x \cdot \frac{1}{x} \right]_1^2$. Evaluating the definite integral: $= \left( e^2 \cdot \frac{1}{2} \right) - \left( e^1 \cdot \frac{1}{1} \right) = \frac{e^2}{2} - e$.

$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx = } $

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