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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
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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