The integral _1^e . ( x e ) ^ 2x - ( e x ) ^x _e ,x ,dx . is equa… - Vidyadip

MCQ

The integral $\int\limits_1^e {\left\{ {\left. {{{\left( {\frac{x}{e}} \right)}^{2x}} - {{\left( {\frac{e}{x}} \right)}^x}} \right\}{{\log }_e}\,x\,dx} \right.} $ is equal to

A
$\frac{1}{2} - e - \frac{1}{{{e^2}}}$
B
$ - \frac{1}{2} + \frac{1}{e} - \frac{1}{{2{e^2}}}$
C
$\frac{3}{2} - \frac{1}{e} - \frac{1}{{2{e^2}}}$
✓
$\frac{3}{2} - e - \frac{1}{{2{e^2}}}$

✓

Answer

Correct option: D.

$\frac{3}{2} - e - \frac{1}{{2{e^2}}}$

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

$\operatorname{let}\left(\frac{x}{e}\right)^{2 x}=t,\left(\frac{e}{x}\right)^{x}=v$

$=\frac{1}{2} \int_{\left(\frac{1}{e}\right)^{2}}^{1} d t+\int_{e}^{1} d v=\frac{1}{2}\left(1-\frac{1}{e^{2}}\right)+(1-e)=\frac{3}{2}-\frac{1}{2 e^{2}}-e$

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