Question

Write a value of\[\int\frac{1}{1 + 2 e^x} \text{ dx }\].

Answer 1

\[\text{ Let I} = \int\frac{dx}{1 + 2 e^x}\]
\[\text{Dividing numerator and denominator by e}^x \]
\[ \Rightarrow I = \int\frac{\frac{1}{e^x}dx}{\frac{1}{e^x} + 2}\]
\[ = \int\frac{e^{- x} dx}{e^{- x} + 2}\]
\[\text{ Let e}^{- x} + 2 = t\]
\[ \Rightarrow - e^{- x}\text{  dx }= dt\]
\[ \Rightarrow e^{- x} \text{ dx }= - dt\]
\[ \therefore I = - \int\frac{dt}{t}\]
\[ = - \text{ log }\left| t \right| + C\]
\[ = - \text{ log }\left| e^{- x} + 2 \right| + C \left( \because t = e^{- x} + 2 \right)\]