Integrate the function frac{e^{2 x}-e^{-2 x}} | vedclass

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Question

Let $I = \int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} dx$. Substitute $t = e^{2 x}+e^{-2 x}$. Differentiating both sides with respect to $x$,we get:

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Vedclass · Answer

Let $I = \int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} dx$.
Substitute $t = e^{2 x}+e^{-2 x}$.
Differentiating both sides with respect to $x$,we get:
$dt = (2e^{2 x} - 2e^{-2 x}) dx = 2(e^{2 x} - e^{-2 x}) dx$.
Therefore,$(e^{2 x} - e^{-2 x}) dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int \frac{1}{t} dt$.
Integrating,we get:
$I = \frac{1}{2} \log |t| + C$.
Substituting back $t = e^{2 x}+e^{-2 x}$:
$I = \frac{1}{2} \log |e^{2 x}+e^{-2 x}| + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$.

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