`int_(0)^(1) (dx)/(e^(x)+e^(-x))`

Explanation

To solve the integral

\[ I = \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}}, \]

we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand:

\[ e^{x} + e^{-x} = \frac{e^{2x} + 1}{e^{x}}. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{1} \frac{dx}{\frac{e^{2x} + 1}{e^{x}}} = \int_{0}^{1} \frac{e^{x} \, dx}{e^{2x} + 1}. \]

### Step 2: Substitution
Next, we will use the substitution \( t = e^{x} \). Then, we have:

\[ dt = e^{x} \, dx \quad \Rightarrow \quad dx = \frac{dt}{t}. \]

When \( x = 0 \), \( t = e^{0} = 1 \). When \( x = 1 \), \( t = e^{1} = e \). Thus, the limits of integration change from \( 0 \) to \( 1 \) to \( 1 \) to \( e \).

Substituting these into the integral gives:

\[ I = \int_{1}^{e} \frac{1}{t^{2} + 1} \, dt. \]

### Step 3: Evaluate the integral
We know that the integral

\[ \int \frac{1}{t^{2} + 1} \, dt = \tan^{-1}(t) + C. \]

Thus, we can evaluate:

\[ I = \left[ \tan^{-1}(t) \right]_{1}^{e} = \tan^{-1}(e) - \tan^{-1}(1). \]

### Step 4: Simplify the result
We know that

\[ \tan^{-1}(1) = \frac{\pi}{4}. \]

Therefore, the result becomes:

\[ I = \tan^{-1}(e) - \frac{\pi}{4}. \]

### Final Result
Thus, the value of the integral is:

\[ I = \tan^{-1}(e) - \frac{\pi}{4}. \]

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