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

Answer

Let's solve the given integrals step by step.

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

1. **Rewrite the integrand**: We can use partial fraction decomposition to express the integrand in a simpler form. We want to find constants \(A\) and \(B\) such that:
\[ \frac{1}{(x+1)(2x+1)} = \frac{A}{x+1} + \frac{B}{2x+1} \]
Multiplying through by the denominator \((x+1)(2x+1)\) gives:
\[ 1 = A(2x + 1) + B(x + 1) \]

2. **Expand and collect terms**:
\[ 1 = (2A + B)x + (A + B) \]
This gives us the system of equations:
\[ 2A + B = 0 \quad \text{(1)} \]
\[ A + B = 1 \quad \text{(2)} \]

3. **Solve the system of equations**:
From equation (1), we can express \(B\) in terms of \(A\):
\[ B = -2A \]
Substituting into equation (2):
\[ A - 2A = 1 \implies -A = 1 \implies A = -1 \]
Then substituting back to find \(B\):
\[ B = -2(-1) = 2 \]

4. **Rewrite the integral**:
Thus, we have:
\[ \frac{1}{(x+1)(2x+1)} = \frac{-1}{x+1} + \frac{2}{2x+1} \]
Therefore, the integral becomes:
\[ \int \frac{dx}{(x+1)(2x+1)} = \int \left( \frac{-1}{x+1} + \frac{2}{2x+1} \right) dx \]

5. **Integrate term by term**:
\[ = -\ln|x+1| + 2 \cdot \frac{1}{2} \ln|2x+1| + C \]
\[ = -\ln|x+1| + \ln|2x+1| + C \]

6. **Combine logarithms**:
\[ = \ln\left|\frac{2x+1}{x+1}\right| + C \]

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

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### Part (ii): \(\int \frac{x-1}{(x-2)(x-3)} dx\)

1. **Rewrite the integrand**: Again, we will use partial fraction decomposition:
\[ \frac{x-1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3} \]
Multiplying through by the denominator \((x-2)(x-3)\) gives:
\[ x - 1 = A(x - 3) + B(x - 2) \]

2. **Expand and collect terms**:
\[ x - 1 = (A + B)x + (-3A - 2B) \]
This gives us the system of equations:
\[ A + B = 1 \quad \text{(1)} \]
\[ -3A - 2B = -1 \quad \text{(2)} \]

3. **Solve the system of equations**:
From equation (1), we can express \(B\) in terms of \(A\):
\[ B = 1 - A \]
Substituting into equation (2):
\[ -3A - 2(1 - A) = -1 \implies -3A - 2 + 2A = -1 \implies -A - 2 = -1 \implies -A = 1 \implies A = -1 \]
Then substituting back to find \(B\):
\[ B = 1 - (-1) = 2 \]

4. **Rewrite the integral**:
Thus, we have:
\[ \frac{x-1}{(x-2)(x-3)} = \frac{-1}{x-2} + \frac{2}{x-3} \]
Therefore, the integral becomes:
\[ \int \frac{x-1}{(x-2)(x-3)} dx = \int \left( \frac{-1}{x-2} + \frac{2}{x-3} \right) dx \]

5. **Integrate term by term**:
\[ = -\ln|x-2| + 2\ln|x-3| + C \]

6. **Combine logarithms**:
\[ = \ln\left|\frac{(x-3)^2}{x-2}\right| + C \]

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

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