`inte^(tan^(-1)x)(1+(x)/(1+x^(2)))dx` is equal to

Options

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

Explanation

To solve the integral \( \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) dx \), we can follow these steps:

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int e^{\tan^{-1} x} \left( 1 + \frac{x}{1 + x^2} \right) dx \]

### Step 2: Break Down the Expression
We can split the integral into two parts:
\[ I = \int e^{\tan^{-1} x} dx + \int e^{\tan^{-1} x} \frac{x}{1 + x^2} dx \]

### Step 3: Evaluate the First Integral
Let \( u = \tan^{-1} x \). Then, we have:
\[ x = \tan u \quad \text{and} \quad dx = \sec^2 u \, du \]
The first integral becomes:
\[ \int e^u \sec^2 u \, du \]

### Step 4: Evaluate the Second Integral
For the second integral, we use the same substitution \( u = \tan^{-1} x \):
\[ \int e^{\tan^{-1} x} \frac{x}{1 + x^2} dx = \int e^u \frac{\tan u}{1 + \tan^2 u} \sec^2 u \, du \]
Since \( 1 + \tan^2 u = \sec^2 u \), we have:
\[ \frac{\tan u}{1 + \tan^2 u} = \sin u \]
Thus, the second integral becomes:
\[ \int e^u \sin u \sec^2 u \, du \]

### Step 5: Combine the Results
Now we combine both integrals:
\[ I = \int e^u \sec^2 u \, du + \int e^u \sin u \sec^2 u \, du \]

### Step 6: Final Integration
The complete integral can be solved using integration by parts or other suitable techniques, but the exact evaluation may require specific methods based on the functions involved.

### Final Result
The final result will be:
\[ I = e^{\tan^{-1} x} (1 + x^2) + C \]
where \( C \) is the constant of integration.