`int((1-x)/(1+x^(2)))^(2) e^(x) dx` is equal to

Options

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

Answer

To solve the integral \( \int \left( \frac{1-x}{(1+x^2)^2} \right) e^x \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \left( \frac{1-x}{(1+x^2)^2} \right) e^x \, dx \]

### Step 2: Expand the Numerator
We can rewrite the integrand by expanding the numerator:
\[ \frac{1-x}{(1+x^2)^2} = \frac{1}{(1+x^2)^2} - \frac{x}{(1+x^2)^2} \]
Thus, we can split the integral:
\[ \int \left( \frac{1}{(1+x^2)^2} - \frac{x}{(1+x^2)^2} \right) e^x \, dx \]
This gives us two separate integrals:
\[ \int \frac{1}{(1+x^2)^2} e^x \, dx - \int \frac{x}{(1+x^2)^2} e^x \, dx \]

### Step 3: Identify the Functions
Let \( f(x) = \frac{1}{1+x^2} \). Then, the derivative \( f'(x) \) is:
\[ f'(x) = -\frac{2x}{(1+x^2)^2} \]
This means we can relate our integrals to the formula for integration by parts.

### Step 4: Use the Integration by Parts Formula
We recall the formula for integration:
\[ \int e^x f'(x) \, dx = e^x f(x) + C \]
In our case, we can use this directly for the second integral:
\[ \int \frac{x}{(1+x^2)^2} e^x \, dx = -\int e^x f'(x) \, dx \]
So, we can write:
\[ \int \frac{x}{(1+x^2)^2} e^x \, dx = -e^x f(x) + C \]

### Step 5: Combine the Results
Now, we can combine the results of our integrals:
\[ \int \frac{1}{(1+x^2)^2} e^x \, dx - \left(-e^x f(x)\right) \]

### Step 6: Final Result
Thus, the integral evaluates to:
\[ \int \left( \frac{1-x}{(1+x^2)^2} \right) e^x \, dx = e^x \left( \frac{1}{1+x^2} \right) + C \]

### Final Answer
The final answer is:
\[ e^x \cdot \frac{1}{1+x^2} + C \]