Tính \(\displaystyle \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^{2x}} - 1}}dx} \)
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Lời giải:
Báo saiTa có: \(\displaystyle \frac{{{e^x}}}{{{e^{2x}} - 1}} = \frac{{{e^x}}}{{\left( {{e^x} - 1} \right)\left( {{e^x} + 1} \right)}}\) \(\displaystyle = \frac{1}{2}\left( {\frac{{{e^x}}}{{{e^x} - 1}} - \frac{{{e^x}}}{{{e^x} + 1}}} \right)\)
Khi đó \(\displaystyle \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^{2x}} - 1}}dx} \) \(\displaystyle = \frac{1}{2}\int\limits_{\frac{1}{2}}^1 {\left( {\frac{{{e^x}}}{{{e^x} - 1}} - \frac{{{e^x}}}{{{e^x} + 1}}} \right)dx} \) \(\displaystyle = \frac{1}{2}\left[ {\int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^x} - 1}}dx} - \int\limits_{\frac{1}{2}}^1 {\frac{{{e^x}}}{{{e^x} + 1}}} dx} \right]\)
\(\displaystyle = \frac{1}{2}\left[ {\int\limits_{\frac{1}{2}}^1 {\frac{{d\left( {{e^x}} \right)}}{{{e^x} - 1}}} - \int\limits_{\frac{1}{2}}^1 {\frac{{d\left( {{e^x}} \right)}}{{{e^x} + 1}}} } \right]\) \(\displaystyle = \frac{1}{2}\left. {\left[ {\ln \left| {{e^x} - 1} \right| - \ln \left| {{e^x} + 1} \right|} \right]} \right|_{\frac{1}{2}}^1\) \(\displaystyle = \frac{1}{2}\left. {\left[ {\ln \left| {\frac{{{e^x} - 1}}{{{e^x} + 1}}} \right|} \right]} \right|_{\frac{1}{2}}^1\)
\(\displaystyle = \frac{1}{2}\left( {\ln \frac{{e - 1}}{{e + 1}} - \ln \frac{{\sqrt e - 1}}{{\sqrt e + 1}}} \right)\) \(\displaystyle = \frac{1}{2}\ln \frac{{\left( {e - 1} \right)\left( {\sqrt e + 1} \right)}}{{\left( {e + 1} \right)\left( {\sqrt e - 1} \right)}}\).
