# #int(2e^x - 2e^-x) / (e^x+e^-x)^2 dx =# ?

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To solve the integral (\int \frac{{2e^x - 2e^{-x}}}{{(e^x+e^{-x})^2}} , dx), we can start by letting (u = e^x + e^{-x}). Then, (du = (e^x - e^{-x}) , dx). Substituting these into the integral, we have:

[ \begin{align*} \int \frac{{2e^x - 2e^{-x}}}{{(e^x+e^{-x})^2}} , dx &= \int \frac{{2(e^x - e^{-x})}}{{(e^x+e^{-x})^2}} , dx \ &= \int \frac{{2du}}{{u^2}} \ &= -\frac{2}{u} + C \ &= -\frac{2}{{e^x + e^{-x}}} + C. \end{align*} ]

So, the solution to the integral is (-\frac{2}{{e^x + e^{-x}}} + C), where (C) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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