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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{dx}}{{1 + {e^x}}} = } $
  • A
    $\log (1 + {e^x})$
  • ✓
    $ - \log (1 + {e^{ - x}})$
  • C
    $ - \log (1 - {e^{ - x}})$
  • D
    $\log ({e^{ - x}} + {e^{ - 2x}})$
✓

Answer

Correct option: B.
$ - \log (1 + {e^{ - x}})$
b
(b)$\int_{}^{} {\frac{{dx}}{{1 + {e^x}}}} = \int_{}^{} {\frac{{{e^{ - x}}}}{{1 + {e^{ - x}}}}} \,dx$
Put $1 + {e^{ - x}} = t$ ==> ${e^{ - x}}dx = - dt$, then it reduces to
$ - \int {\frac{{dt}}{t} = - \log t = - \log (1 + {e^{ - x}})} $.

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