Read the following passages and answer the following questions (7-9)
Consider the integrals of the form `l=inte^(x)(f(x)+f'(x))dx` By product rule considering `e^(x)f(x)` as first integral and `e^(x)f'(x)` as second one, we get `l=e^(x)f(x)-int(f(x)+f'(x))dx=e^(x)f(x)+c`
`l=inte^(x)(tan^(-1)x+(2x)/((1+x^(2))^(2)))dx` then l is equal to

Options

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

Answer

To solve the integral \( l = \int e^x \left( \tan^{-1} x + \frac{2x}{(1+x^2)^2} \right) dx \), we will apply the product rule and the properties of integration as discussed in the passage.

### Step-by-Step Solution:

1. **Identify the Function**:
We can express the integral in a form that allows us to apply the product rule. Let:
\[ f(x) = \tan^{-1} x - \frac{1}{1+x^2} \]
The derivative \( f'(x) \) will help us simplify the integral.

2. **Calculate the Derivative**:
To find \( f'(x) \):
\[ f'(x) = \frac{d}{dx} \left( \tan^{-1} x \right) - \frac{d}{dx} \left( \frac{1}{1+x^2} \right) \]
The derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \), and the derivative of \( \frac{1}{1+x^2} \) is:
\[ -\frac{2x}{(1+x^2)^2} \]
Therefore:
\[ f'(x) = \frac{1}{1+x^2} + \frac{2x}{(1+x^2)^2} \]

3. **Combine \( f(x) \) and \( f'(x) \)**:
Now we can see that:
\[ f'(x) = \frac{1}{1+x^2} + \frac{2x}{(1+x^2)^2} \]
This means:
\[ \tan^{-1} x + f'(x) = \tan^{-1} x + \left( \frac{1}{1+x^2} + \frac{2x}{(1+x^2)^2} \right) \]

4. **Apply the Product Rule**:
According to the product rule, we have:
\[ \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \]
Thus:
\[ l = e^x f(x) + C \]

5. **Substitute \( f(x) \)**:
Substitute \( f(x) \) back into the equation:
\[ l = e^x \left( \tan^{-1} x - \frac{1}{1+x^2} \right) + C \]

6. **Final Result**:
Therefore, the value of \( l \) is:
\[ l = e^x \left( \tan^{-1} x - \frac{1}{1+x^2} \right) + C \]