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

Answer

Let's solve the three integrals step by step.

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

**Step 1: Substitution**
Let \( t = e^x - 1 \). Then, \( e^x = t + 1 \) and \( dx = \frac{dt}{e^x} = \frac{dt}{t + 1} \).

**Step 2: Rewrite the integral**
Substituting into the integral, we have:
\[ \int \frac{e^x}{e^x(e^x - 1)} \, dx = \int \frac{t + 1}{(t + 1)t} \cdot \frac{dt}{t + 1} = \int \frac{1}{t} - \frac{1}{t + 1} \, dt \]

**Step 3: Integrate**
Now, we can integrate:
\[ \int \left( \frac{1}{t} - \frac{1}{t + 1} \right) dt = \ln |t| - \ln |t + 1| + C \]

**Step 4: Substitute back**
Substituting back \( t = e^x - 1 \):
\[ \ln |e^x - 1| - \ln |e^x| + C = \ln \left( \frac{e^x - 1}{e^x} \right) + C = \ln \left( 1 - \frac{1}{e^x} \right) + C \]

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

---

### Part (ii):
Evaluate the integral:
\[ \int \frac{e^x}{(1 + e^x)(2 + e^x)} \, dx \]

**Step 1: Substitution**
Let \( t = e^x + 1 \). Then, \( e^x = t - 1 \) and \( dx = \frac{dt}{e^x} = \frac{dt}{t - 1} \).

**Step 2: Rewrite the integral**
Substituting into the integral, we have:
\[ \int \frac{t - 1}{(t)(t + 1)} \cdot \frac{dt}{t - 1} = \int \frac{1}{t} - \frac{1}{t + 1} \, dt \]

**Step 3: Integrate**
Now, we can integrate:
\[ \int \left( \frac{1}{t} - \frac{1}{t + 1} \right) dt = \ln |t| - \ln |t + 1| + C \]

**Step 4: Substitute back**
Substituting back \( t = e^x + 1 \):
\[ \ln |e^x + 1| - \ln |e^x + 2| + C = \ln \left( \frac{e^x + 1}{e^x + 2} \right) + C \]

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

---

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

**Step 1: Substitution**
Let \( t = e^x - 1 \). Then, \( e^x = t + 1 \) and \( dx = \frac{dt}{t + 1} \).

**Step 2: Rewrite the integral**
Substituting into the integral, we have:
\[ \int \frac{t + 1}{t^2(t + 3)} \cdot \frac{dt}{t + 1} = \int \frac{1}{t^2} - \frac{1}{t + 3} \, dt \]

**Step 3: Integrate**
Now, we can integrate:
\[ \int \left( \frac{1}{t^2} - \frac{1}{t + 3} \right) dt = -\frac{1}{t} - \ln |t + 3| + C \]

**Step 4: Substitute back**
Substituting back \( t = e^x - 1 \):
\[ -\frac{1}{e^x - 1} - \ln |e^x + 2| + C \]

### Final Answer for Part (iii):
\[ \int \frac{e^x}{(e^x - 1)^2(e^x + 2)} \, dx = -\frac{1}{e^x - 1} - \ln |e^x + 2| + C \]

---