Practicing Success
Practicing Success
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| The value of the integration \(\int \frac{dx}{(1+e^{x})(1-e^{-x})}\) is |
\(\frac{e^{2}}{1+e^{2}}\) \(-\frac{e^{2}}{1+e^{2}}\) \(\frac{1}{1+e^{2}}\) \(-\frac{1}{1+e^{2}}\) |
| \(-\frac{1}{1+e^{2}}\) |
| Let \(1+e^{2}=t\) so \(e^{x}dx=dt\) \(\begin{aligned}\int \frac{dx}{(1+e^{x})(1-e^{-x})}&=\int \frac{e^{x}}{(1+e^{x})^{2}}dx\\ &=\frac{-1}{1+e^{2}}\end{aligned}\) |
