Evaluate :
`int_( (dx)/(e^(x)+e^(-x)) )`

Answer

To evaluate the integral

\[ I = \int \frac{dx}{e^x + e^{-x}}, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the denominator:
\[ e^x + e^{-x} = \frac{e^{2x} + 1}{e^x}. \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{e^x + e^{-x}} = \int \frac{e^x \, dx}{e^{2x} + 1}. \]

### Step 2: Substitution
Now, we will use the substitution:
\[ t = e^x \implies dt = e^x \, dx \implies dx = \frac{dt}{t}. \]
Substituting this into the integral gives:
\[ I = \int \frac{dt}{t^2 + 1}. \]

### Step 3: Recognize the Standard Integral
The integral \(\int \frac{dt}{t^2 + 1}\) is a standard integral that evaluates to:
\[ \int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + C. \]

### Step 4: Substitute Back
Now, we substitute back \(t = e^x\):
\[ I = \tan^{-1}(e^x) + C. \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{dx}{e^x + e^{-x}} = \tan^{-1}(e^x) + C. \]

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