The value of the integral `inte^(x)((1-x)/(1+x^(2)))^(2)dx` is

Options

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

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the integrand
We start by rewriting the integrand:

\[ \left( \frac{1-x}{1+x^2} \right)^2 = \frac{(1-x)^2}{(1+x^2)^2}. \]

Thus, the integral becomes:

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

### Step 2: Expand the numerator
Next, we expand the numerator \((1-x)^2\):

\[ (1-x)^2 = 1 - 2x + x^2. \]

So, we can rewrite the integral as:

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

### Step 3: Solve each integral separately
We will solve the integrals one by one.

1. **First Integral:**
\[ I_1 = \int e^x \frac{1}{(1+x^2)^2} dx. \]

2. **Second Integral:**
\[ I_2 = -2 \int e^x \frac{x}{(1+x^2)^2} dx. \]

3. **Third Integral:**
\[ I_3 = \int e^x \frac{x^2}{(1+x^2)^2} dx. \]

### Step 4: Apply integration by parts
For \(I_2\) and \(I_3\), we will apply integration by parts.

For \(I_2\):
Let \(u = \frac{1}{(1+x^2)^2}\) and \(dv = e^x dx\).
Then \(du = -\frac{4x}{(1+x^2)^3} dx\) and \(v = e^x\).

Using integration by parts:
\[ I_2 = \left[ \frac{e^x}{(1+x^2)^2} \right] - \int e^x \left(-\frac{4x}{(1+x^2)^3}\right) dx. \]

For \(I_3\):
Let \(u = \frac{x^2}{(1+x^2)^2}\) and \(dv = e^x dx\).
Then \(du = \left(\frac{2x(1+x^2) - 2x^3}{(1+x^2)^4}\right) dx\) and \(v = e^x\).

Using integration by parts:
\[ I_3 = \left[ \frac{x^2 e^x}{(1+x^2)^2} \right] - \int e^x \left(\frac{2x(1+x^2) - 2x^3}{(1+x^2)^4}\right) dx. \]

### Step 5: Combine results
After calculating \(I_1\), \(I_2\), and \(I_3\), we combine them to get the final result.

### Final Answer
The value of the integral

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

where \(C\) is the constant of integration.

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