Evaluate : `int x^n log x dx.`

Answer

To evaluate the integral \( I = \int x^n \log x \, dx \), we will use the integration by parts formula. The integration by parts formula states:

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

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \log x \) (which means \( du = \frac{1}{x} \, dx \))
- \( dv = x^n \, dx \) (which means \( v = \frac{x^{n+1}}{n+1} \))

### Step 2: Apply the integration by parts formula
Using the integration by parts formula, we have:

\[ I = \int x^n \log x \, dx = uv - \int v \, du \]

Substituting our choices for \( u \), \( du \), \( v \), and \( dv \):

\[ I = \log x \cdot \frac{x^{n+1}}{n+1} - \int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx \]

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

\[ I = \frac{x^{n+1} \log x}{n+1} - \frac{1}{n+1} \int x^n \, dx \]

### Step 4: Evaluate the remaining integral
Now, we need to evaluate \( \int x^n \, dx \):

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

### Step 5: Substitute back into the equation
Substituting this back into our expression for \( I \):

\[ I = \frac{x^{n+1} \log x}{n+1} - \frac{1}{n+1} \left( \frac{x^{n+1}}{n+1} + C \right) \]

### Step 6: Final simplification
Now, we can combine the terms:

\[ I = \frac{x^{n+1} \log x}{n+1} - \frac{x^{n+1}}{(n+1)^2} - \frac{C}{n+1} \]

Thus, we can write the final answer as:

\[ I = \frac{x^{n+1} \log x}{n+1} - \frac{x^{n+1}}{(n+1)^2} + C \]

### Final Answer
\[ \int x^n \log x \, dx = \frac{x^{n+1} \log x}{n+1} - \frac{x^{n+1}}{(n+1)^2} + C \]