`inte^(x){tan^(-1)x+(1)/((1+x^(2)))}dx=?`

Options

A. `e^(x)*(1)/((1+x^(2)))+C`
B. `e^(x)tan^(-1)x+C`
C. `-e^(x)cot^(-1)x+C`
D. none of these

Explanation

To solve the integral \( \int e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right) dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Let the integral be denoted as \( I \)**:
\[ I = \int e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right) dx \]

2. **Use substitution**:
We will use the substitution:
\[ t = e^x \tan^{-1} x \]
To differentiate \( t \) with respect to \( x \), we will apply the product rule.

3. **Differentiate \( t \)**:
Using the product rule:
\[ \frac{dt}{dx} = e^x \tan^{-1} x + e^x \cdot \frac{1}{1 + x^2} \]
This gives us:
\[ \frac{dt}{dx} = e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right) \]

4. **Rearranging for \( dx \)**:
From the above equation, we can express \( dx \):
\[ dx = \frac{dt}{e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right)} \]

5. **Substituting back into the integral**:
Substitute \( dx \) back into the integral:
\[ I = \int e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right) dx = \int dt \]

6. **Integrate**:
The integral of \( dt \) is simply:
\[ I = t + C \]

7. **Substituting back for \( t \)**:
Recall that \( t = e^x \tan^{-1} x \), so:
\[ I = e^x \tan^{-1} x + C \]

### Final Answer:
\[ \int e^x \left( \tan^{-1} x + \frac{1}{1 + x^2} \right) dx = e^x \tan^{-1} x + C \]