Evaluate the integrals : `I = int_(0)^(1) (e^(x))/( 1 + e^(2x)) dx` ,

Answer

To evaluate the integral

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

we can use a substitution method. Let's proceed step by step.

### Step 1: Substitution
Let \( T = e^x \). Then, the differential \( dT = e^x \, dx \) or \( dx = \frac{dT}{T} \).

### Step 2: Change the limits of integration
When \( x = 0 \), \( T = e^0 = 1 \).
When \( x = 1 \), \( T = e^1 = e \).

Thus, the limits of integration change from \( x = 0 \) to \( x = 1 \) into \( T = 1 \) to \( T = e \).

### Step 3: Rewrite the integral
Now substituting \( e^x \) and \( dx \) in the integral, we have:

\[ I = \int_{1}^{e} \frac{T}{1 + T^2} \cdot \frac{dT}{T} = \int_{1}^{e} \frac{1}{1 + T^2} \, dT. \]

### Step 4: Evaluate the integral
The integral \( \int \frac{1}{1 + T^2} \, dT \) is known to be \( \tan^{-1}(T) \). Thus,

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

### Step 5: Calculate the values
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \). Therefore,

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

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

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

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