`int (2x)/(x^2+3x+2) dx`
(ii) `int x/((x+1)(x+2)) dx`

Answer

Let's solve the given integrals step by step.

### Part (i): \(\int \frac{2x}{x^2 + 3x + 2} \, dx\)

1. **Factor the Denominator**:
The denominator \(x^2 + 3x + 2\) can be factored as:
\[ x^2 + 3x + 2 = (x + 1)(x + 2) \]
So we rewrite the integral:
\[ \int \frac{2x}{(x + 1)(x + 2)} \, dx \]

2. **Rewrite the Numerator**:
We can express \(2x\) as \(2(x + 1) - 2\):
\[ \int \frac{2(x + 1) - 2}{(x + 1)(x + 2)} \, dx = \int \left( \frac{2}{x + 2} - \frac{2}{(x + 1)(x + 2)} \right) \, dx \]

3. **Split the Integral**:
Now we can split the integral into two parts:
\[ \int \frac{2}{x + 2} \, dx - 2 \int \frac{1}{(x + 1)(x + 2)} \, dx \]

4. **Integrate the First Part**:
The first integral is straightforward:
\[ \int \frac{2}{x + 2} \, dx = 2 \ln |x + 2| + C_1 \]

5. **Integrate the Second Part**:
For the second integral, we can use partial fraction decomposition:
\[ \frac{1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \]
Solving for \(A\) and \(B\), we find \(A = 1\) and \(B = -1\):
\[ \int \left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) \, dx = \ln |x + 1| - \ln |x + 2| + C_2 \]

6. **Combine Results**:
Putting it all together:
\[ 2 \ln |x + 2| - 2 \left( \ln |x + 1| - \ln |x + 2| \right) = 2 \ln |x + 2| - 2 \ln |x + 1| + 2 \ln |x + 2| \]
This simplifies to:
\[ 4 \ln |x + 2| - 2 \ln |x + 1| + C \]

### Final Answer for Part (i):
\[ \int \frac{2x}{x^2 + 3x + 2} \, dx = 4 \ln |x + 2| - 2 \ln |x + 1| + C \]

---

### Part (ii): \(\int \frac{x}{(x + 1)(x + 2)} \, dx\)

1. **Use Partial Fraction Decomposition**:
We can express \(\frac{x}{(x + 1)(x + 2)}\) as:
\[ \frac{x}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} \]
Multiplying through by the denominator and solving gives \(A = 1\) and \(B = -1\):
\[ \int \left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) \, dx \]

2. **Integrate**:
Now we can integrate:
\[ \int \frac{1}{x + 1} \, dx - \int \frac{1}{x + 2} \, dx = \ln |x + 1| - \ln |x + 2| + C \]

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

---