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

Options

A. `tan^(-1)e-(pi)/(4)`
B. `tan^(-1)e+(pi)/(4)`
C. `tan^(-1)e`
D. `(pi)/(4)-tan^(-1)e`

Answer

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

### Step 1: Rewrite the integrand
The integrand can be rewritten using the property of exponents:
\[ \frac{1}{e^{x} + e^{-x}} = \frac{1}{\frac{e^{2x} + 1}{e^{x}}} = \frac{e^{x}}{e^{2x} + 1} \]
Thus, we can rewrite the integral as:
\[ \int_{0}^{1} \frac{e^{x}}{e^{2x} + 1} \, dx \]

### Step 2: Substitution
Let \( t = e^{x} \). Then, the differential \( dt = e^{x} \, dx \) or \( dx = \frac{dt}{t} \).
Next, we need to change the limits of integration:
- When \( x = 0 \), \( t = e^{0} = 1 \)
- When \( x = 1 \), \( t = e^{1} = e \)

Now, substituting these into the integral, we get:
\[ \int_{1}^{e} \frac{1}{t^{2} + 1} \, dt \]

### Step 3: Evaluate the integral
The integral \( \int \frac{1}{t^{2} + 1} \, dt \) is a standard integral that evaluates to \( \tan^{-1}(t) \). Therefore, we have:
\[ \int_{1}^{e} \frac{1}{t^{2} + 1} \, dt = \tan^{-1}(t) \Big|_{1}^{e} \]
Calculating this gives:
\[ \tan^{-1}(e) - \tan^{-1}(1) \]

### Step 4: Substitute values
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \). Thus, the result becomes:
\[ \tan^{-1}(e) - \frac{\pi}{4} \]

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

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