_1^2 e^ 2x ( 1 x - 1 2 x^2 ) ,dx is equal to - Vidyadip

MCQ

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

A
$\frac{{{e^2}}}{4}\left( {{e^2} + 2} \right)$
B
${e^2}\left( {{e^2} - 2} \right)$
C
$\frac{{{e^2}\left( {{e^2} - 2} \right)}}{2}$
✓
$\frac{{{e^2}\left( {{e^2} - 2} \right)}}{4}$

✓

Answer

Correct option: D.

$\frac{{{e^2}\left( {{e^2} - 2} \right)}}{4}$

d
${\rm{I}} = \int_1^2 {{\rm{e}}_{II}^{2x}} \cdot \mathop {\frac{1}{{\rm{x}}}}\limits_I {\rm{dx}} - \int_1^2 {{{\rm{e}}^{2{\rm{x}}}}} \cdot \frac{{\ln {\rm{x}}}}{{2{{\rm{x}}^2}}}{\rm{dx}}$

$=\left(\frac{1}{\mathrm{x}} \cdot \frac{\mathrm{e}^{2 \mathrm{x}}}{2}\right)_{1}^{2}-\int_{1}^{2}-\frac{1}{\mathrm{x}^{2}} \cdot \frac{\mathrm{e}^{2 \mathrm{x}}}{2} \mathrm{dx}-\int_{1}^{2} \mathrm{e}^{2 \mathrm{x}} \cdot \frac{1}{2 \mathrm{x}^{2}} \mathrm{d} \mathrm{x}$

$I=\left(\frac{e^{4}}{4}-\frac{e^{2}}{2}\right)=\frac{e^{2}\left(e^{2}-2\right)}{4}$

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