(i) `int x^2 log x dx`
(ii) `int (x^2+1) log x dx`

Answer

To solve the integrals given in the question, we will use integration by parts. The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]

where \( u \) is a function we choose to differentiate, and \( dv \) is a function we choose to integrate.

### Part (i): Evaluate \( \int x^2 \log x \, dx \)

1. **Choose \( u \) and \( dv \)**:
- Let \( u = \log x \) (which we will differentiate)
- Let \( dv = x^2 \, dx \) (which we will integrate)

2. **Differentiate and Integrate**:
- Then, \( du = \frac{1}{x} \, dx \)
- And, \( v = \int x^2 \, dx = \frac{x^3}{3} \)

3. **Apply Integration by Parts**:
\[ \int x^2 \log x \, dx = uv - \int v \, du \]
Substituting the values we have:
\[ = \log x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \]
Simplifying the integral:
\[ = \log x \cdot \frac{x^3}{3} - \frac{1}{3} \int x^2 \, dx \]

4. **Evaluate the Remaining Integral**:
\[ = \log x \cdot \frac{x^3}{3} - \frac{1}{3} \cdot \frac{x^3}{3} + C \]
\[ = \log x \cdot \frac{x^3}{3} - \frac{x^3}{9} + C \]

Thus, the final result for part (i) is:
\[ \int x^2 \log x \, dx = \frac{x^3}{3} \log x - \frac{x^3}{9} + C \]

### Part (ii): Evaluate \( \int (x^2 + 1) \log x \, dx \)

1. **Split the Integral**:
\[ \int (x^2 + 1) \log x \, dx = \int x^2 \log x \, dx + \int \log x \, dx \]

2. **Evaluate \( \int \log x \, dx \)** using integration by parts:
- Let \( u = \log x \) and \( dv = dx \)
- Then, \( du = \frac{1}{x} \, dx \) and \( v = x \)

3. **Apply Integration by Parts**:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx \]
\[ = x \log x - \int 1 \, dx = x \log x - x + C \]

4. **Combine Results**:
Now, substituting back into the expression for part (ii):
\[ \int (x^2 + 1) \log x \, dx = \left( \frac{x^3}{3} \log x - \frac{x^3}{9} \right) + \left( x \log x - x \right) + C \]
Combine the terms:
\[ = \frac{x^3}{3} \log x + x \log x - \frac{x^3}{9} - x + C \]

Thus, the final result for part (ii) is:
\[ \int (x^2 + 1) \log x \, dx = \left( \frac{x^3}{3} + x \right) \log x - \frac{x^3}{9} - x + C \]