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

Explanation

To solve the integral \( \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}} \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression in the denominator:
\[ e^{x} + e^{-x} = \frac{e^{2x} + 1}{e^{x}} \]
Thus, we can rewrite the integral as:
\[ \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}} = \int_{0}^{1} \frac{e^{x}}{e^{2x} + 1} \, dx \]

### Step 2: Substitute \( t = e^{x} \)
Now, we perform the substitution \( t = e^{x} \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{dt}{t} \]
When \( x = 0 \), \( t = e^{0} = 1 \), and when \( x = 1 \), \( t = e^{1} = e \). Therefore, the limits of integration change from \( 0 \) to \( 1 \) into \( 1 \) to \( e \).

### Step 3: Rewrite the integral in terms of \( t \)
Substituting \( t \) into the integral, we have:
\[ \int_{1}^{e} \frac{1}{t^2 + 1} \, dt \]

### Step 4: Evaluate the integral
The integral \( \int \frac{1}{t^2 + 1} \, dt \) is known to be:
\[ \tan^{-1}(t) \]
Thus, we can evaluate the definite integral:
\[ \left[ \tan^{-1}(t) \right]_{1}^{e} = \tan^{-1}(e) - \tan^{-1}(1) \]

### Step 5: Substitute the values
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \), so we have:
\[ \tan^{-1}(e) - \frac{\pi}{4} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}} = \tan^{-1}(e) - \frac{\pi}{4} \]