If `int x^(2) ln x dx = (x^(3))/(m) ln x + (x^(3))/(n) + c`, then what are the values of m and n respectively ?
where c is a constant od intergration

Options

A. `(1)/(3) - (1)/(9)`
B. `3, -9`
C. `3, 9`
D. `3, 3`

Explanation

To solve the integral \( \int x^2 \ln x \, dx \) and find the values of \( m \) and \( n \) in the expression

\[ \int x^2 \ln x \, dx = \frac{x^3}{m} \ln x + \frac{x^3}{n} + c, \]

we can use integration by parts.

### Step 1: Set up integration by parts
Let:
- \( u = \ln x \) (then \( du = \frac{1}{x} \, dx \))
- \( dv = x^2 \, dx \) (then \( v = \frac{x^3}{3} \))

### Step 2: Apply integration by parts formula
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ \int x^2 \ln x \, dx = \left( \ln x \cdot \frac{x^3}{3} \right) - \int \left( \frac{x^3}{3} \cdot \frac{1}{x} \right) dx \]

### Step 3: Simplify the integral
The integral simplifies to:

\[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \]

\[ = \frac{x^3}{3} \ln x - \frac{1}{3} \int x^2 \, dx \]

### Step 4: Solve the remaining integral
Now, compute \( \int x^2 \, dx \):

\[ \int x^2 \, dx = \frac{x^3}{3} + C \]

Substituting this back, we get:

\[ \int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \frac{1}{3} \left( \frac{x^3}{3} \right) + C \]

\[ = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \]

### Step 5: Compare with the given expression
Now we compare this with the expression given in the question:

\[ \int x^2 \ln x \, dx = \frac{x^3}{m} \ln x + \frac{x^3}{n} + c \]

From our result, we have:

\[ \frac{x^3}{3} \ln x - \frac{x^3}{9} + C \]

This implies:

- \( \frac{1}{m} = \frac{1}{3} \) → \( m = 3 \)
- \( \frac{1}{n} = -\frac{1}{9} \) → \( n = -9 \)

### Final Answer
Thus, the values of \( m \) and \( n \) are:

\[ m = 3, \quad n = -9 \]