$\int\limits_1^2 {\left( {\cfrac{1}{x} - \cfrac{1}{{2{x^2}}}} \right){e^{2x}}dx}$

: Let $I = \int\limits_1^2 {{e^{2x}}\left( {\cfrac{1}{x} - \cfrac{1}{{2{x^2}}}} \right)dx}$

Put $2x = t$ $\Rightarrow$ $2dx = dt$

When $x = 1,t = 2$ and when $x = 2,t = 4$

$\therefore$ $I = \cfrac{1}{2}\int\limits_2^4 {{e^t}\left( {\cfrac{2}{t} - \cfrac{{1 \times 4}}{{2{t^2}}}} \right)} dt = \cfrac{1}{2}\int\limits_2^4 {{e^t}\left( {\cfrac{2}{t} - \cfrac{2}{{{t^2}}}} \right)dt}$

$= \int\limits_2^4 {{e^t} \cdot \left( {\cfrac{1}{t} - \cfrac{1}{{{t^2}}}} \right)dt}$

$= \int\limits_2^4 {{e^t} \cdot \left[ {\cfrac{1}{t} + \cfrac{d}{{dt}}\left( {\cfrac{1}{t}} \right)} \right]dt}$
$= \left[ {{e^t} \cdot \cfrac{1}{t}} \right]_2^4 = \cfrac{1}{4}{e^4} - \cfrac{{{e^2}}}{2} = \cfrac{{{e^2}}}{2}\left( {\cfrac{{{e^2}}}{2} - 1} \right) = \cfrac{{{e^2}\left( {{e^2} - 2} \right)}}{4}$

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