What is integral `int ` 1/(e^x + e^(-x)) dx?

### 1 Answer | Add Yours

You need to re-write the denominator at integrand, using the negative exponent property, such that:

`int 1/(e^x + e^(-x))dx = int 1/(e^x + 1/e^x)dx`

`e^x + 1/e^x = (e^(2x) + 1)/e^x`

Replacing `(e^(2x) + 1)/e^x` for the original form of denominator, yields:

`int 1/(e^x + e^(-x))dx = int 1/((e^(2x) + 1)/e^x) dx`

`int 1/(e^x + e^(-x))dx = int e^x/(e^(2x) + 1) dx`

You need to come up with the substitution `e^x = y` , such that:

`e^x = y => e^x dx = dy`

`int e^x/(e^(2x) + 1) dx = int 1/(y^2 + 1) dy`

`int 1/(y^2 + 1) dy = tan^(-1) y + c`

Replacing back `e^x` for `y` yields:

`int 1/(e^x + e^(-x))dx = tan^(-1) (e^x) + c `

c represents an aleatory real constant

**Hence, evaluating the indefinite integral, using suggested substitution, yields **`int 1/(e^x + e^(-x))dx = tan^(-1) (e^x) + c .`

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes