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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{e^x + e^{-x}} \, dx \]
We can rewrite \( e^{-x} \) as \( \frac{1}{e^x} \):
\[ I = \int \frac{1}{e^x + \frac{1}{e^x}} \, dx = \int \frac{1}{\frac{e^{2x} + 1}{e^x}} \, dx \]
This simplifies to:
\[ I = \int \frac{e^x}{e^{2x} + 1} \, dx \]

### Step 2: Substitution
Now, we will use the substitution \( t = e^x \). Then, the differential \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \). Substituting these into the integral gives:
\[ I = \int \frac{t}{t^2 + 1} \cdot \frac{dt}{t} = \int \frac{1}{t^2 + 1} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{t^2 + 1} \, dt \) is a standard integral that evaluates to:
\[ I = \tan^{-1}(t) + C \]
where \( C \) is the constant of integration.

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

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

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