Evaluate: `int1/(e^x+e^(-x))\ dx`

Explanation

To evaluate the integral \( \int \frac{1}{e^x + e^{-x}} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{1}{e^x + e^{-x}} = \frac{1}{\frac{e^{2x} + 1}{e^x}} = \frac{e^x}{e^{2x} + 1} \]

### Step 2: Substitute
Let \( t = e^x \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{dt}{t} \]
Now, substituting \( e^x = t \) into the integral gives:
\[ \int \frac{e^x}{e^{2x} + 1} \, dx = \int \frac{t}{t^2 + 1} \cdot \frac{dt}{t} = \int \frac{1}{t^2 + 1} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{t^2 + 1} \, dt \) is a standard integral, which results in:
\[ \int \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) + C \]

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

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

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