intlimits_1^2 {{e^{2x}}} left( {frac{1}{x} - | vedclass

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(D) Let $I = \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx$. We know the standard integral form $\int e^{ax} [f(x) + \frac{f'(x)}{a}]

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$\frac{{{e^4}}}{4} - \frac{{{e^2}}}{2}$

Vedclass · Answer

(D) Let $I = \int_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx$. We know the standard integral form $\int e^{ax} [f(x) + \frac{f'(x)}{a}] dx = \frac{1}{a} e^{ax} f(x) + C$. Here,let $f(x) = \frac{1}{x}$. Then $f'(x) = -\frac{1}{x^2}$. Comparing with the given integral,$a = 2$. Thus,$\int e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) dx = \frac{1}{2} e^{2x} \left( \frac{1}{x} \right) = \frac{e^{2x}}{2x}$. Now,applying the limits from $1$ to $2$: $I = \left[ \frac{e^{2x}}{2x} \right]_1^2 = \frac{e^{2(2)}}{2(2)} - \frac{e^{2(1)}}{2(1)} = \frac{e^4}{4} - \frac{e^2}{2}$. This can be written as $\frac{e^4 - 2e^2}{4}$.

$\int\limits_1^2 {{e^{2x}}} \left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right)\,dx$ is equal to

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