What is the result of the integral ∫ (e^(2x) + e^(-2x)) dx?

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To solve the integral ∫ (e^(2x) + e^(-2x)) dx, we can address the integral by breaking it down into two separate integrals.

First, consider the integral of e^(2x):

∫ e^(2x) dx = (1/2)e^(2x) + C₁.

We reach this result by applying the formula ∫ e^(ax) dx = (1/a)e^(ax) + C, where a is 2 in this case.

Next, we consider the integral of e^(-2x):

∫ e^(-2x) dx = (-1/2)e^(-2x) + C₂.

Here, we use the same formula but with a = -2, yielding a negative outcome.

Combining both parts, we have:

∫ (e^(2x) + e^(-2x)) dx = ∫ e^(2x) dx + ∫ e^(-2x) dx = (1/2)e^(2x) - (1/2)e^(-2x) + C.

Thus, the result of the integral is indeed (1/2)e^(2x) - (1