(i) `int_2^3 1/x dx`
(ii) `int_1^2 1/x^2 dx`
(iii) `int_1^e (dx)/x`
(iv) `int_-1^1 (x+1) dx`

Answer

Let's solve the given integrals step by step:

### (i) \(\int_2^3 \frac{1}{x} \, dx\)

1. **Identify the integral**: We need to integrate \(\frac{1}{x}\) from 2 to 3.
2. **Find the antiderivative**: The antiderivative of \(\frac{1}{x}\) is \(\ln |x|\).
3. **Apply the limits**: We evaluate \(\ln |x|\) from 2 to 3:
\[ \int_2^3 \frac{1}{x} \, dx = \left[ \ln x \right]_2^3 = \ln 3 - \ln 2 \]
4. **Use logarithmic properties**: We can simplify this using the property \(\ln a - \ln b = \ln \frac{a}{b}\):
\[ \ln 3 - \ln 2 = \ln \frac{3}{2} \]

### Final answer for (i):
\[ \int_2^3 \frac{1}{x} \, dx = \ln \frac{3}{2} \]

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### (ii) \(\int_1^2 \frac{1}{x^2} \, dx\)

1. **Identify the integral**: We need to integrate \(\frac{1}{x^2}\) from 1 to 2.
2. **Find the antiderivative**: The antiderivative of \(\frac{1}{x^2}\) is \(-\frac{1}{x}\).
3. **Apply the limits**: We evaluate \(-\frac{1}{x}\) from 1 to 2:
\[ \int_1^2 \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_1^2 = -\frac{1}{2} - (-1) = -\frac{1}{2} + 1 = \frac{1}{2} \]

### Final answer for (ii):
\[ \int_1^2 \frac{1}{x^2} \, dx = \frac{1}{2} \]

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### (iii) \(\int_1^e \frac{1}{x} \, dx\)

1. **Identify the integral**: We need to integrate \(\frac{1}{x}\) from 1 to \(e\).
2. **Find the antiderivative**: The antiderivative of \(\frac{1}{x}\) is \(\ln |x|\).
3. **Apply the limits**: We evaluate \(\ln x\) from 1 to \(e\):
\[ \int_1^e \frac{1}{x} \, dx = \left[ \ln x \right]_1^e = \ln e - \ln 1 \]
4. **Simplify**: Since \(\ln e = 1\) and \(\ln 1 = 0\):
\[ \ln e - \ln 1 = 1 - 0 = 1 \]

### Final answer for (iii):
\[ \int_1^e \frac{1}{x} \, dx = 1 \]

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### (iv) \(\int_{-1}^1 (x + 1) \, dx\)

1. **Identify the integral**: We need to integrate \(x + 1\) from -1 to 1.
2. **Find the antiderivative**: The antiderivative of \(x + 1\) is \(\frac{x^2}{2} + x\).
3. **Apply the limits**: We evaluate \(\frac{x^2}{2} + x\) from -1 to 1:
\[ \int_{-1}^1 (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{-1}^1 \]
4. **Calculate at the limits**:
- At \(x = 1\):
\[ \frac{1^2}{2} + 1 = \frac{1}{2} + 1 = \frac{3}{2} \]
- At \(x = -1\):
\[ \frac{(-1)^2}{2} + (-1) = \frac{1}{2} - 1 = -\frac{1}{2} \]
5. **Combine the results**:
\[ \frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \]

### Final answer for (iv):
\[ \int_{-1}^1 (x + 1) \, dx = 2 \]

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### Summary of Answers:
1. \(\int_2^3 \frac{1}{x} \, dx = \ln \frac{3}{2}\)
2. \(\int_1^2 \frac{1}{x^2} \, dx = \frac{1}{2}\)
3. \(\int_1^e \frac{1}{x} \, dx = 1\)
4. \(\int_{-1}^1 (x + 1) \, dx = 2\)

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