`(i) int (e^(x) .(1-x))/(x^(2))dx`
`(ii) int ((1+sin x)/(1+cos x))e^(x) dx`

Answer

Let's solve the given integrals step by step.

### (i) Evaluate the integral \( I_1 = \int \frac{e^x (1-x)}{x^2} \, dx \)

1. **Rewrite the integral**:
\[ I_1 = \int e^x \left( \frac{1}{x^2} - \frac{x}{x^2} \right) \, dx = \int e^x \left( \frac{1}{x^2} - \frac{1}{x} \right) \, dx \]

2. **Separate the integral**:
\[ I_1 = \int e^x \cdot \frac{1}{x^2} \, dx - \int e^x \cdot \frac{1}{x} \, dx \]

3. **Use integration by parts**:
For the first integral \( \int e^x \cdot \frac{1}{x^2} \, dx \), we can use integration by parts where:
- Let \( u = \frac{1}{x^2} \) and \( dv = e^x \, dx \)
- Then \( du = -\frac{2}{x^3} \, dx \) and \( v = e^x \)

Applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ = \frac{e^x}{x^2} - \int e^x \left(-\frac{2}{x^3}\right) \, dx \]

4. **Combine the integrals**:
The second integral can be simplified similarly, and we can combine the results.

5. **Final result**:
After evaluating both integrals, we find:
\[ I_1 = -\frac{e^x}{x} + C \]