(i) `int_0^1 (dx)/sqrt(1-x^2)`
(ii) `int_0^1 (dx)/sqrt(1+x^2)`
(iii) a `int_1^sqrt3 (dx)/(1+x^2)`
b `int_0^1 (dx)/(1+x^2)`
(iv) `int_0^(2//3) (dx)/(4+9x^2)`
(v) `int_0^1 x/(x^2+1)dx`
(vi) `int_2^3 x/(x^2+1) dx`

Answer

Let's solve the integrals step by step.

### (i) \(\int_0^1 \frac{dx}{\sqrt{1-x^2}}\)

1. **Recognize the Integral**: The integral \(\int \frac{dx}{\sqrt{1-x^2}}\) is a standard integral that evaluates to \(\sin^{-1}(x) + C\).
2. **Evaluate the Definite Integral**:
\[ \int_0^1 \frac{dx}{\sqrt{1-x^2}} = \left[\sin^{-1}(x)\right]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) \]
3. **Substitute the Limits**:
\[ = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### (ii) \(\int_0^1 \frac{dx}{\sqrt{1+x^2}}\)

1. **Recognize the Integral**: The integral \(\int \frac{dx}{\sqrt{1+x^2}}\) evaluates to \(\ln(x + \sqrt{1+x^2}) + C\).
2. **Evaluate the Definite Integral**:
\[ \int_0^1 \frac{dx}{\sqrt{1+x^2}} = \left[\ln(x + \sqrt{1+x^2})\right]_0^1 \]
3. **Substitute the Limits**:
\[ = \ln(1 + \sqrt{2}) - \ln(0 + 1) = \ln(1 + \sqrt{2}) - 0 = \ln(1 + \sqrt{2}) \]

### (iii) (a) \(\int_1^{\sqrt{3}} \frac{dx}{1+x^2}\)

1. **Recognize the Integral**: The integral \(\int \frac{dx}{1+x^2}\) evaluates to \(\tan^{-1}(x) + C\).
2. **Evaluate the Definite Integral**:
\[ \int_1^{\sqrt{3}} \frac{dx}{1+x^2} = \left[\tan^{-1}(x)\right]_1^{\sqrt{3}} \]
3. **Substitute the Limits**:
\[ = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) = \frac{\pi}{3} - \frac{\pi}{4} \]
4. **Simplify**:
\[ = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

### (iii) (b) \(\int_0^1 \frac{dx}{1+x^2}\)

1. **Evaluate the Definite Integral**:
\[ \int_0^1 \frac{dx}{1+x^2} = \left[\tan^{-1}(x)\right]_0^1 \]
2. **Substitute the Limits**:
\[ = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### (iv) \(\int_0^{\frac{2}{3}} \frac{dx}{4+9x^2}\)

1. **Rewrite the Integral**:
\[ = \int_0^{\frac{2}{3}} \frac{dx}{4(1 + \frac{9}{4}x^2)} = \frac{1}{4} \int_0^{\frac{2}{3}} \frac{dx}{1 + \left(\frac{3}{2}x\right)^2} \]
2. **Recognize the Integral**: The integral \(\int \frac{dx}{1 + k^2 x^2}\) evaluates to \(\frac{1}{k} \tan^{-1}(kx) + C\).
3. **Evaluate**:
\[ = \frac{1}{4} \cdot \frac{2}{3} \tan^{-1}\left(\frac{3}{2}x\right) \bigg|_0^{\frac{2}{3}} = \frac{1}{6} \left[\tan^{-1}(1) - \tan^{-1}(0)\right] \]
4. **Substitute the Limits**:
\[ = \frac{1}{6} \left(\frac{\pi}{4} - 0\right) = \frac{\pi}{24} \]

### (v) \(\int_0^1 \frac{x}{x^2+1}dx\)

1. **Rewrite the Integral**:
\[ = \frac{1}{2} \int_0^1 \frac{2x}{x^2+1}dx \]
2. **Recognize the Integral**: The integral \(\int \frac{2x}{x^2+1}dx\) evaluates to \(\ln(x^2 + 1) + C\).
3. **Evaluate**:
\[ = \frac{1}{2} \left[\ln(x^2 + 1)\right]_0^1 = \frac{1}{2} \left(\ln(2) - \ln(1)\right) = \frac{1}{2} \ln(2) \]

### (vi) \(\int_2^3 \frac{x}{x^2+1}dx\)

1. **Rewrite the Integral**:
\[ = \frac{1}{2} \int_2^3 \frac{2x}{x^2+1}dx \]
2. **Evaluate**:
\[ = \frac{1}{2} \left[\ln(x^2 + 1)\right]_2^3 = \frac{1}{2} \left(\ln(10) - \ln(5)\right) \]
3. **Simplify**:
\[ = \frac{1}{2} \ln\left(\frac{10}{5}\right) = \frac{1}{2} \ln(2) \]

### Summary of Results:
1. (i) \(\frac{\pi}{2}\)
2. (ii) \(\ln(1+\sqrt{2})\)
3. (iii) (a) \(\frac{\pi}{12}\), (b) \(\frac{\pi}{4}\)
4. (iv) \(\frac{\pi}{24}\)
5. (v) \(\frac{1}{2} \ln(2)\)
6. (vi) \(\frac{1}{2} \ln(2)\)