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

Answer

To solve the integral \( \int \left( \frac{1-x}{(1+x^2)^2} \right) e^x \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ I = \int \left( \frac{1-x}{(1+x^2)^2} \right) e^x \, dx \]
We can separate the fraction:
\[ I = \int \left( \frac{1}{(1+x^2)^2} - \frac{x}{(1+x^2)^2} \right) e^x \, dx \]

2. **Split the Integral**:
This allows us to split the integral into two parts:
\[ I = \int \frac{e^x}{(1+x^2)^2} \, dx - \int \frac{x e^x}{(1+x^2)^2} \, dx \]

3. **Evaluate the First Integral**:
Let’s denote:
\[ I_1 = \int \frac{e^x}{(1+x^2)^2} \, dx \]
and
\[ I_2 = \int \frac{x e^x}{(1+x^2)^2} \, dx \]

4. **Integration by Parts for \( I_2 \)**:
We will use integration by parts for \( I_2 \):
\[ u = \frac{1}{1+x^2}, \quad dv = x e^x \, dx \]
Then, we find:
\[ du = -\frac{2x}{(1+x^2)^2} \, dx, \quad v = e^x \]
Applying integration by parts:
\[ I_2 = uv - \int v \, du \]
This gives:
\[ I_2 = \frac{e^x}{1+x^2} - \int e^x \left(-\frac{2x}{(1+x^2)^2}\right) \, dx \]

5. **Combine the Integrals**:
Now we can substitute \( I_2 \) back into our expression for \( I \):
\[ I = I_1 - \left( \frac{e^x}{1+x^2} - \int \frac{2x e^x}{(1+x^2)^2} \, dx \right) \]

6. **Simplifying the Expression**:
Notice that the integral \( \int \frac{2x e^x}{(1+x^2)^2} \, dx \) appears on both sides of the equation. This allows us to cancel out the terms:
\[ I + \int \frac{2x e^x}{(1+x^2)^2} \, dx = I_1 + \frac{e^x}{1+x^2} \]
Thus, we can simplify to find:
\[ 2I = I_1 + \frac{e^x}{1+x^2} \]
Therefore:
\[ I = \frac{1}{2} \left( I_1 + \frac{e^x}{1+x^2} \right) \]

7. **Final Result**:
After evaluating \( I_1 \) and combining the results, we find:
\[ I = \frac{e^x}{1+x^2} + C \]

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