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$\int_{-5}^{5}\left[\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right] d x=$
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$\begin{aligned}
\text { Let } f(x)=& \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \\
=& \frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1} \text { and } f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}=\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}} \\
\therefore f(x)=-f(x)
\end{aligned}$
Thus $f(x)$ is an odd function.
$\therefore \int_{-5}^{5} \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} d x=0$
\text { Let } f(x)=& \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \\
=& \frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1} \text { and } f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}=\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}} \\
\therefore f(x)=-f(x)
\end{aligned}$
Thus $f(x)$ is an odd function.
$\therefore \int_{-5}^{5} \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} d x=0$
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