If `f(x)=inte^(x)(tan^(-1)x+(2x)/((1+x^(2))^(2)))dx,f(0)=0` then the value of f(1) is

Options

A. `e((pi)/(4)-(1)/(2))+1`
B. `e((pi)/(4)+(1)/(2))+1`
C. `e((pi)/(2)-(1)/(4))+1`
D. `e^(-1)((pi)/(4)-(1)/(2))+1`

Explanation

To solve the problem, we need to evaluate the function defined by the integral:

\[ f(x) = \int e^x \left( \tan^{-1}(x) + \frac{2x}{(1+x^2)^2} \right) dx \]

with the condition that \( f(0) = 0 \). We are tasked with finding \( f(1) \).

### Step-by-Step Solution:

1. **Rewrite the Integral:**
We can separate the integral into two parts:
\[ f(x) = \int e^x \tan^{-1}(x) \, dx + \int e^x \frac{2x}{(1+x^2)^2} \, dx \]

2. **Use Integration by Parts:**
For the first integral, we can use integration by parts. Let:
- \( u = \tan^{-1}(x) \) ⇒ \( du = \frac{1}{1+x^2} \, dx \)
- \( dv = e^x \, dx \) ⇒ \( v = e^x \)

Applying integration by parts:
\[ \int e^x \tan^{-1}(x) \, dx = e^x \tan^{-1}(x) - \int e^x \frac{1}{1+x^2} \, dx \]

3. **Evaluate the Second Integral:**
For the second integral:
\[ \int e^x \frac{2x}{(1+x^2)^2} \, dx \]
We can notice that:
\[ \frac{d}{dx} \left( \frac{1}{1+x^2} \right) = -\frac{2x}{(1+x^2)^2} \]
Thus, we can rewrite the integral as:
\[ \int e^x \frac{2x}{(1+x^2)^2} \, dx = -\int e^x \frac{d}{dx} \left( \frac{1}{1+x^2} \right) \, dx \]

Using integration by parts again, we can find:
\[ -\int e^x \frac{d}{dx} \left( \frac{1}{1+x^2} \right) \, dx = -e^x \frac{1}{1+x^2} + \int e^x \frac{1}{1+x^2} \, dx \]

4. **Combine Results:**
Combining all parts, we have:
\[ f(x) = e^x \tan^{-1}(x) - \left( e^x \frac{1}{1+x^2} - \int e^x \frac{1}{1+x^2} \, dx \right) + C \]

5. **Substituting \( f(0) = 0 \):**
Now we need to find the constant \( C \) using the condition \( f(0) = 0 \):
\[ f(0) = e^0 \tan^{-1}(0) - \left( e^0 \frac{1}{1+0^2} - \int e^0 \frac{1}{1+0^2} \, dx \right) + C = 0 \]
Simplifying:
\[ 0 - (1 - 0) + C = 0 \implies C = 1 \]

6. **Finding \( f(1) \):**
Now we can find \( f(1) \):
\[ f(1) = e^1 \tan^{-1}(1) - \left( e^1 \frac{1}{1+1^2} - \int e^1 \frac{1}{1+1^2} \, dx \right) + 1 \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \):
\[ f(1) = e \cdot \frac{\pi}{4} - \left( e \cdot \frac{1}{2} - \frac{e}{2} \right) + 1 \]
Simplifying gives:
\[ f(1) = e \cdot \frac{\pi}{4} - \frac{e}{2} + 1 \]

### Final Answer:
Thus, the value of \( f(1) \) is:
\[ f(1) = e \cdot \frac{\pi}{4} - \frac{e}{2} + 1 \]