Sum
Write a value of\[\int\frac{1}{1 + e^x} \text{ dx }\]
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Solution
\[\text{ Let I }= \int\frac{dx}{1 + e^x}\]
\[\text{ Dividing numerator and denominator by e}^x \]
\[ \Rightarrow I = \int\frac{\frac{1}{e^x}\text{ dx}}{\frac{1}{e^x} + 1}\]
\[ = \int\frac{e^{- x}\text{ dx}}{e^{- x} + 1}\]
\[\text{ Let e}^{- x} + 1 = t\]
\[ - e^{- x} dx = dt\]
\[ \Rightarrow e^{- x} dx = - dt\]
\[ \therefore I = \int - \frac{dt}{t}\]
\[ = - \text{ log }\left| t \right| + C\]
\[ = - \text{ log }\left| 1 + e^x \right| + C \left( \because t = 1 + e^x \right)\]
Concept: Methods of Integration: Integration by Substitution
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