What is `int_(0)^(1) (e^(tan^(-1))x dx)/(1+x^(2)) ` equal to

Options

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

Explanation

To solve the integral

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

we can use a substitution method.

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, the derivative of \( t \) with respect to \( x \) is

\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \implies dx = (1 + x^2) dt. \]

### Step 2: Change the limits
When \( x = 0 \), \( t = \tan^{-1}(0) = 0 \).
When \( x = 1 \), \( t = \tan^{-1}(1) = \frac{\pi}{4} \).

### Step 3: Substitute in the integral
Now we substitute \( x \) and \( dx \) in the integral:

\[ I = \int_{0}^{\frac{\pi}{4}} e^t \, dt. \]

### Step 4: Integrate
The integral of \( e^t \) is \( e^t \). Thus, we have:

\[ I = \left[ e^t \right]_{0}^{\frac{\pi}{4}} = e^{\frac{\pi}{4}} - e^0 = e^{\frac{\pi}{4}} - 1. \]

### Final Answer
Therefore, the value of the integral is

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

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