What is the integral of (1 + e^(-x))^2 dx?

What is the integral of (1 + e^(-x))^2 dx?

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

Answer: (C)

To find the integral of \( (1 + e^{-x})^2 \, dx \), we can start by expanding the integrand: \[ (1 + e^{-x})^2 = 1 + 2e^{-x} + e^{-2x} \] Now we can integrate each term separately: 1. The integral of \( 1 \) with respect to \( x \) is \( x \). 2. The integral of \( 2e^{-x} \) can be found using the formula for the integral of the exponential function. The integral is: \[ \int 2e^{-x} \, dx = -2e^{-x} \] 3. Finally, for \( e^{-2x} \): \[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \] Putting all of this together, we have: \[ \int (1 + e^{-x})^2 \, dx = \int (1 + 2e^{-x} + e^{-2x}) \, dx = x - 2e^{-x} - \frac{1}{

To find the integral of ( (1 + e^{-x})^2 , dx ), we can start by expanding the integrand:

[

(1 + e^{-x})^2 = 1 + 2e^{-x} + e^{-2x}

]

Now we can integrate each term separately:

1. The integral of ( 1 ) with respect to ( x ) is ( x ).

2. The integral of ( 2e^{-x} ) can be found using the formula for the integral of the exponential function. The integral is:

[

\int 2e^{-x} , dx = -2e^{-x}

]

3. Finally, for ( e^{-2x} ):

[

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

]

Putting all of this together, we have:

[

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