What is the integral of e^(2x) + e^(-2x) with respect to x?

What is the integral of e^(2x) + e^(-2x) with respect to x?

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

Answer: (A)

To find the integral of \( e^{2x} + e^{-2x} \) with respect to \( x \), we can integrate each term separately. The integral of \( e^{2x} \) is computed using the formula \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \). Here, \( a = 2 \), so: \[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C_1 \] Similarly, for \( e^{-2x} \), we again use the integration formula where \( a = -2 \): \[ \int e^{-2x} \, dx = \frac{1}{-2} e^{-2x} + C_2 = -\frac{1}{2} e^{-2x} + C_2 \] Putting these results together, the integral of \( e^{2x} + e^{-2x} \) becomes: \[ \int (e^{2x} + e^{-2x}) \, dx = \left( \frac{1}{2} e

To find the integral of ( e^{2x} + e^{-2x} ) with respect to ( x ), we can integrate each term separately.

The integral of ( e^{2x} ) is computed using the formula ( \int e^{ax} , dx = \frac{1}{a} e^{ax} + C ). Here, ( a = 2 ), so:

[

\int e^{2x} , dx = \frac{1}{2} e^{2x} + C_1

]

Similarly, for ( e^{-2x} ), we again use the integration formula where ( a = -2 ):

[

\int e^{-2x} , dx = \frac{1}{-2} e^{-2x} + C_2 = -\frac{1}{2} e^{-2x} + C_2

]

Putting these results together, the integral of ( e^{2x} + e^{-2x} ) becomes:

[

\int (e^{2x} + e^{-2x}) , dx = \left( \frac{1}{2} e