`(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.

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

**Step 1: Split the integral**
We can separate the integrand:
\[ I_1 = \int \left( \frac{e^x}{x^2} - \frac{e^x x}{x^2} \right) \, dx = \int \frac{e^x}{x^2} \, dx - \int \frac{e^x}{x} \, dx \]

**Step 2: Recognize the forms**
The integrals \(\int \frac{e^x}{x^2} \, dx\) and \(\int \frac{e^x}{x} \, dx\) can be solved using integration by parts or known results.

**Step 3: Integration by parts**
Let’s use integration by parts for both integrals. For the first integral, let:
- \(u = \frac{1}{x}\) and \(dv = e^x dx\)
- Then \(du = -\frac{1}{x^2} dx\) and \(v = e^x\)

Using integration by parts:
\[ \int \frac{e^x}{x^2} \, dx = \frac{e^x}{x} - \int e^x \left(-\frac{1}{x^2}\right) dx = \frac{e^x}{x} + \int \frac{e^x}{x^2} \, dx \]
This leads to a recursive relation.

**Step 4: Solve the recursive relation**
We can denote:
\[ I = \int \frac{e^x}{x^2} \, dx \]
Then:
\[ I = \frac{e^x}{x} + I \]
This implies:
\[ I = \frac{e^x}{x} + C \]

**Step 5: Combine results**
Thus:
\[ I_1 = \left( \frac{e^x}{x} + C_1 \right) - \left( e^x \ln |x| + C_2 \right) \]
The final result is:
\[ I_1 = \frac{e^x}{x} - e^x \ln |x| + C \]

### Part (ii):
Evaluate the integral
\[ I_2 = \int \frac{1 + \sin x}{1 + \cos x} e^x \, dx \]

**Step 1: Simplify the integrand**
We can rewrite:
\[ \frac{1 + \sin x}{1 + \cos x} = \frac{(1 + \cos x) + \sin x - \cos x}{1 + \cos x} = 1 + \frac{\sin x - \cos x}{1 + \cos x} \]
Thus:
\[ I_2 = \int e^x \, dx + \int \frac{(\sin x - \cos x)e^x}{1 + \cos x} \, dx \]

**Step 2: Solve the first integral**
The first integral is straightforward:
\[ \int e^x \, dx = e^x + C \]

**Step 3: Solve the second integral**
For the second part, we can use integration by parts again. Let:
- \(u = \frac{\sin x - \cos x}{1 + \cos x}\) and \(dv = e^x dx\)
- Then \(du\) and \(v\) can be calculated accordingly.

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]

**Step 4: Combine results**
After solving, we will have:
\[ I_2 = e^x + \text{(result from the second integral)} + C \]

### Final Answers:
1. \(I_1 = \frac{e^x}{x} - e^x \ln |x| + C\)
2. \(I_2 = e^x + \text{(result from the second integral)} + C\)