Correct Answer - Option 2 : ln(1 + e
x) + C
Concept:
1. Integration by Substitution:
- If the given integration is of the form \(\smallint {\rm{g}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right){\rm{f'}}\left( {\rm{x}} \right){\rm{dx}}\) where \({\rm{g}}\left( {\rm{x}} \right)\) and \({\rm{f}}\left( {\rm{x}} \right)\) are both differentiable functions then we substitute \({\rm{f}}\left( {\rm{x}} \right) = {\rm{u}}\) which implies that \({\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{du}}\).
- Therefore, the integral becomes \(\smallint {\rm{g}}\left( {\rm{u}} \right){\rm{du}}\) which can be solved by general formulas.
Solution:
First simplify the given integral by writing the denominator as \({{\rm{e}}^{ - {\rm{x}}}} = {\rm{}}\frac{1}{{{{\rm{e}}^{\rm{x}}}}}\).
Therefore, the integral becomes:
\(\smallint \frac{{{\rm{dx}}}}{{1 + {{\rm{e}}^{ - {\rm{x}}}}}} = \smallint \frac{{{{\rm{e}}^{\rm{x}}}}}{{1 + {{\rm{e}}^{\rm{x}}}}}{\rm{dx}}\)
Now substitute \(1 + {{\rm{e}}^{\rm{x}}} = {\rm{u}}\) therefore \({{\rm{e}}^{\rm{x}}}{\rm{dx}} = {\rm{du}}\).
The integral becomes,
\(\smallint \frac{{{\rm{du}}}}{{\rm{u}}} = \ln \left( {\rm{u}} \right) + C\)
Resubstitute \({\rm{u}} = 1 + {{\rm{e}}^{\rm{x}}}\).
Therefore,
\(\smallint \frac{{{\rm{dx}}}}{{1 + {{\rm{e}}^{ - {\rm{x}}}}}} = \ln \left( {1 + {{\rm{e}}^{\rm{x}}}} \right) + C\).
