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

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

• A. (1/2)e^(2x) - (1/2)e^(-2x) + C
• B. (1/4)e^(2x) - (1/4)e^(-2x) + C
• C. e^(2x) + e^(-2x) + C
• D. (1/3)e^(2x) + (1/3)e^(-2x) + C

Answer: (A)

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

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