Evaluate the integral of dx/1+e^x from -1 to 0
∫−10dx1+ex=∫−101+ex−ex1+exdx=∫−101+ex1+exdx−∫−10ex1+exdx==∫−10dx−∫−10d(1+ex)1+ex=x∣−10−ln(1+ex)∣−10==0+1−ln(1+1)+ln(1+e−1)=1−ln2+ln(e+1e)==1+ln(e+12e)\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}−1∫01+exdx=−1∫01+ex1+ex−exdx=−1∫01+ex1+exdx−−1∫01+exexdx==−1∫0dx−−1∫01+exd(1+ex)=x∣−10−ln(1+ex)∣−10==0+1−ln(1+1)+ln(1+e−1)=1−ln2+ln(ee+1)==1+ln(2ee+1)
Answer: 1+ln(e+12e)1 + \ln \left( {\frac{{e + 1}}{{2e}}} \right)1+ln(2ee+1)
