The value of the definite integral, _1^ ( e^ x + 1 + e^ 3 - x ) ^… - Vidyadip

MCQ

The value of the definite integral,$\int\limits_1^\infty {{{({e^{x + 1}} + {e^{3 - x}})}^{ - 1}}\,dx} $ is

✓
$\frac{\pi }{{4{e^2}}}$
B
$\frac{\pi }{{4e}}$
C
$\frac{1}{{{e^2}}}\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}\frac{1}{e}} \right)$
D
$\frac{\pi }{{2{e^2}}}$

✓

Answer

Correct option: A.

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

a
$I$ =$\int\limits_1^\infty {\frac{{dx}}{{(e\,\cdot\,{e^x} + {e^3}\,\cdot\,{e^{ - x}})}}} $ =$\int\limits_1^\infty {\frac{{{e^x}\,dx}}{{e({e^{2x}} + {e^2})}}} $ (multiply $N^r $ and $D^r $ by $e^x$ )
put $e^x = t$ $\Rightarrow $ $e^x dx = dt$
$I =$ $\frac{1}{e}\,\,\int\limits_e^\infty {\frac{{dt}}{{{t^2} + {e^2}}}} $= $\left. {\frac{1}{{{e^2}}}{{\tan }^{ - 1}}\frac{t}{e}} \right|_e^\infty $ =$\frac{1}{{{e^2}}}\left[ {\frac{\pi }{2} - \frac{\pi }{4}} \right]$ = $\frac{\pi }{{4{e^2}}}$

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