Integrals — Class 12 Maths Solution

ncert misc SA NCERT Misc.,Q.41,Page.353
Question

$\int {\cfrac{{dx}}{{{e^x} + {e^{ - x}}}}}$is equal to

  • (a) ${\tan ^{ - 1}}\left( {{e^x}} \right) + C$
  • (b) ${\tan ^{ - 1}}\left( {{e^{ - x}}} \right) + C$
  • (c) $\log \left( {{e^x} - {e^{ - x}}} \right) + C$
  • (d) $\log \left( {{e^x} + {e^{ - x}}} \right) + C$
Step-by-step Solution

Option a is correct

: Let $I = \int {\cfrac{{dx}}{{{e^x} + {e^{ - x}}}} = } \int {\cfrac{{dx}}{{{e^x} + \cfrac{1}{{{e^x}}}}}}$

$\Rightarrow$ $\int {\cfrac{{{e^x}dx}}{{{{\left( {{e^x}} \right)}^2} + 1}}}$

Put ${e^x} = t$ $\Rightarrow$ ${e^x}dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{{1 + {t^2}}}} = {\tan ^{ - 1}}\left( t \right) + C = {\tan ^{ - 1}}\left( {{e^x}} \right) + C$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.