(i) `int_1^3 (3x)/(9x^2-1) dx`
(ii) `int_2^3 x/(x^2+1) dx`

Answer

Let's solve the integrals step by step.

### Part (i): Evaluate the integral \( \int_1^3 \frac{3x}{9x^2 - 1} \, dx \)

**Step 1: Substitution**
Let \( y = 3x \). Then, we have:
\[ dy = 3 \, dx \quad \Rightarrow \quad dx = \frac{dy}{3} \]
Also, we need to change the limits of integration:
- When \( x = 1 \), \( y = 3 \cdot 1 = 3 \)
- When \( x = 3 \), \( y = 3 \cdot 3 = 9 \)

**Step 2: Rewrite the integral**
Substituting \( y \) and \( dx \) into the integral, we get:
\[ \int_1^3 \frac{3x}{9x^2 - 1} \, dx = \int_3^9 \frac{y}{9\left(\frac{y}{3}\right)^2 - 1} \cdot \frac{dy}{3} \]
This simplifies to:
\[ = \frac{1}{3} \int_3^9 \frac{y}{9 \cdot \frac{y^2}{9} - 1} \, dy = \frac{1}{3} \int_3^9 \frac{y}{y^2 - 1} \, dy \]

**Step 3: Simplify the integral**
Now, we can simplify the integral:
\[ = \frac{1}{3} \int_3^9 \frac{y}{(y-1)(y+1)} \, dy \]

**Step 4: Partial fraction decomposition**
We can express \( \frac{y}{(y-1)(y+1)} \) as:
\[ \frac{y}{(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} \]
Multiplying through by the denominator gives:
\[ y = A(y+1) + B(y-1) \]
Setting \( y = 1 \) gives \( 1 = 2A \) so \( A = \frac{1}{2} \).
Setting \( y = -1 \) gives \( -1 = -2B \) so \( B = \frac{1}{2} \).
Thus,
\[ \frac{y}{(y-1)(y+1)} = \frac{1/2}{y-1} + \frac{1/2}{y+1} \]

**Step 5: Integrate**
Now we can integrate:
\[ \frac{1}{3} \int_3^9 \left( \frac{1/2}{y-1} + \frac{1/2}{y+1} \right) dy = \frac{1}{6} \left[ \ln|y-1| + \ln|y+1| \right]_3^9 \]

**Step 6: Evaluate the limits**
Calculating the limits:
\[ = \frac{1}{6} \left[ \ln(8) + \ln(10) - (\ln(2) + \ln(4)) \right] \]
\[ = \frac{1}{6} \left[ \ln(80) - \ln(8) \right] = \frac{1}{6} \ln(10) \]

### Part (ii): Evaluate the integral \( \int_2^3 \frac{x}{x^2 + 1} \, dx \)

**Step 1: Simplify the integral**
We can factor out a constant:
\[ \int_2^3 \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int_2^3 \frac{2x}{x^2 + 1} \, dx \]

**Step 2: Recognize the derivative**
The derivative of \( x^2 + 1 \) is \( 2x \), so we can integrate directly:
\[ = \frac{1}{2} \ln|x^2 + 1| \bigg|_2^3 \]

**Step 3: Evaluate the limits**
Calculating the limits:
\[ = \frac{1}{2} \left[ \ln(10) - \ln(5) \right] = \frac{1}{2} \ln\left(\frac{10}{5}\right) = \frac{1}{2} \ln(2) \]

### Final Answers:
(i) \( \frac{1}{6} \ln(10) \)

(ii) \( \frac{1}{2} \ln(2) \)

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