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 \( f(x) \) defined as:

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

with the condition that \( f(0) = 0 \). We need to find the value of \( f(1) \).

### Step 1: Rewrite the Integral

We can separate the integral into two parts:

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

### Step 2: Use Integration by Parts

For the first integral, we can use integration by parts. Let:

- \( u = \tan^{-1}(x) \) ⇒ \( du = \frac{1}{1+x^2} \, dx \)
- \( dv = e^x \, dx \) ⇒ \( v = e^x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

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

### Step 3: Evaluate the Second Integral

Now we need to evaluate the second integral:

\[ \int e^x \frac{2x}{(1+x^2)^2} \, dx \]

Notice that \( \frac{2x}{(1+x^2)^2} \) is the derivative of \( -\frac{1}{1+x^2} \). Thus, we can rewrite the integral as:

\[ \int e^x \frac{2x}{(1+x^2)^2} \, dx = -e^x \frac{1}{1+x^2} + C \]

### Step 4: Combine the Results

Combining both parts, we have:

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

### Step 5: Determine the Constant \( C \)

We know that \( f(0) = 0 \):

\[ f(0) = e^0 \tan^{-1}(0) - e^0 \frac{1}{1+0^2} + C = 0 \]

This simplifies to:

\[ 0 = 0 - 1 + C \implies C = 1 \]

### Step 6: Final Expression for \( f(x) \)

Thus, we have:

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

### Step 7: Evaluate \( f(1) \)

Now we can find \( f(1) \):

\[ f(1) = e^1 \tan^{-1}(1) - e^1 \frac{1}{1+1^2} + 1 \]

Calculating each term:

- \( e^1 = e \)
- \( \tan^{-1}(1) = \frac{\pi}{4} \)
- \( \frac{1}{1+1^2} = \frac{1}{2} \)

Substituting these values:

\[ f(1) = e \cdot \frac{\pi}{4} - e \cdot \frac{1}{2} + 1 \]

### Step 8: Simplify

\[ f(1) = e \left( \frac{\pi}{4} - \frac{1}{2} \right) + 1 \]

### Final Answer

Thus, the value of \( f(1) \) is:

\[ f(1) = e \left( \frac{\pi}{4} - \frac{2}{4} \right) + 1 = e \left( \frac{\pi - 2}{4} \right) + 1 \]