Answer to Question #162057 in Calculus for Phyroe

Question

Evaluate the integral of dx/1+e^x from -1 to 0

Answer 1

$\begin{array}{l} \int\limits_{ - 1}^0 {\frac{{dx}}{{1 + {e^x}}}} = \int\limits_{ - 1}^0 {\frac{{1 + {e^x} - {e^x}}}{{1 + {e^x}}}dx} = \int\limits_{ - 1}^0 {\frac{{1 + {e^x}}}{{1 + {e^x}}}dx} - \int\limits_{ - 1}^0 {\frac{{{e^x}}}{{1 + {e^x}}}dx} = \\ = \int\limits_{ - 1}^0 {dx} - \int\limits_{ - 1}^0 {\frac{{d(1 + {e^x})}}{{1 + {e^x}}}} = \left. x \right|_{ - 1}^0 - \ln \left. {(1 + {e^x})} \right|_{ - 1}^0 = \\ = 0 + 1 - \ln (1 + 1) + \ln (1 + {e^{ - 1}}) = 1 - \ln 2 + \ln \left( {\frac{{e + 1}}{e}} \right) = \\ = 1 + \ln \left( {\frac{{e + 1}}{{2e}}} \right) \end{array}$

Answer: $1 + \ln \left( {\frac{{e + 1}}{{2e}}} \right)$