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

Options

A. `(pi)/( 4)`
B. `(pi)/( 2)`
C. `tan^(-1) ((e-1)/( e+1))`
D. `tan^(-1) ((1-e)/( 1+e))`

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the integrand

The expression \( e^{x} + e^{-x} \) can be rewritten using the identity for hyperbolic cosine:

\[ e^{x} + e^{-x} = 2 \cosh(x). \]

Thus, we can rewrite the integral as:

\[ \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}} = \int_{0}^{1} \frac{dx}{2 \cosh(x)} = \frac{1}{2} \int_{0}^{1} \frac{dx}{\cosh(x)}. \]

### Step 2: Use substitution

Next, we can use 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 \( x = 1 \), \( t = e^{1} = e \). The limits of integration change accordingly.

### Step 3: Rewrite the integral

Substituting into the integral, we have:

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

### Step 4: Integrate

The integral

\[ \int \frac{t}{t^2 + 1} dt \]

can be solved using the substitution \( u = t^2 + 1 \), which gives \( du = 2t dt \) or \( dt = \frac{du}{2t} \). Thus, we have:

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

### Step 5: Evaluate the definite integral

Now, we evaluate the definite integral from \( 1 \) to \( e \):

\[ \frac{1}{2} \left[ \frac{1}{2} \ln(t^2 + 1) \right]_{1}^{e} = \frac{1}{4} \left[ \ln(e^2 + 1) - \ln(1^2 + 1) \right] = \frac{1}{4} \left[ \ln(e^2 + 1) - \ln(2) \right]. \]

### Step 6: Simplify the result

This simplifies to:

\[ \frac{1}{4} \ln\left(\frac{e^2 + 1}{2}\right). \]

### Final Answer

Thus, the value of the integral

\[ \int_{0}^{1} \frac{dx}{e^{x} + e^{-x}} = \frac{1}{4} \ln\left(\frac{e^2 + 1}{2}\right). \]