Question

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

Answer 1

Answer:

\dfrac{\ln (e^{2x} + e^{-2x})}{2} + c[/tex]

Step-by-step explanation:

Hi,

We need to evaluate the following integral:

Let $I = \int \dfrac{e^{2x} - e^{-2x} dx}{e^{2x} + e^{-2x}}$

Let  $e^{2x} + e^{-2x} = t$ , then

On differentiating, we get

$2e^{2x} - 2e^{-2x} dx = dt$

On substituting, the above integral I will become

$I = \int \dfrac{dt}{2t}$

$I = \dfrac{\ln t}{2} + c$,

where t = $e^{2x} + e^{-2x}$

Hence, $I = \dfrac{\ln (e^{2x} + e^{-2x})}{2} + c$

Hope, it helps !