Integral of 1/e^x+e^-x dx

1. Integrate term-by-term:

   1. Let $u = - x$.

      Then let $du = - dx$ and substitute $- du$:

      $$\int \left(- e^{u}\right)\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\text{False}$$

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         So, the result is: $- e^{u}$

      Now substitute $u$ back in:

      $$- e^{- x}$$

   1. Let $u = e^{x}$.

      Then let $du = e^{x} dx$ and substitute $du$:

      $$\int \frac{1}{u^{2}}\, du$$

      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

      Now substitute $u$ back in:

      $$- e^{- x}$$

   The result is: $- 2 e^{- x}$

2. Add the constant of integration:

   $$- 2 e^{- x}+ \mathrm{constant}$$

The answer is: