Find the following integral. \[ \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x \] \( \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x=\square \)
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To solve the integral \[ \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx, \] let's rewrite the integrand. Notice that \[ \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \frac{\sinh(x)}{\cosh(x)} = \tanh(x). \] Now we can express the integral in terms of the hyperbolic tangent: \[ \int \tanh(x) \, dx. \] The integral of \(\tanh(x)\) is a well-known result: \[ \int \tanh(x) \, dx = -\ln|\cosh(x)| + C, \] where \( C \) is the constant of integration. Thus, the final answer for the integral is: \[ \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \, dx = -\ln|\cosh(x)| + C. \]