`I= int x^(3) "In" x dx`.

Answer

To solve the integral \( I = \int x^3 \ln x \, dx \), we will use the method of integration by parts. The formula for integration by parts is given by:

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

### Step 1: Choose \( u \) and \( dv \)

Let:
- \( u = \ln x \) (which we will differentiate)
- \( dv = x^3 \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and Integrate \( dv \)

Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{x} \, dx \]
- Integrate \( dv \):
\[ v = \int x^3 \, dx = \frac{x^4}{4} \]

### Step 3: Apply the Integration by Parts Formula

Now we can apply the integration by parts formula:
\[ I = \int x^3 \ln x \, dx = uv - \int v \, du \]
Substituting the values we found:
\[ I = \ln x \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ I = \frac{x^4 \ln x}{4} - \frac{1}{4} \int x^3 \, dx \]

### Step 4: Solve the Remaining Integral

Now, we need to solve the integral \( \int x^3 \, dx \):
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Substituting this back into our equation:
\[ I = \frac{x^4 \ln x}{4} - \frac{1}{4} \cdot \frac{x^4}{4} \]
This simplifies to:
\[ I = \frac{x^4 \ln x}{4} - \frac{x^4}{16} \]

### Step 5: Combine the Terms

Now, we can combine the terms:
\[ I = \frac{x^4}{4} \ln x - \frac{x^4}{16} \]

### Step 6: Final Answer

Thus, the final answer for the integral is:
\[ I = \frac{x^4}{4} \ln x - \frac{x^4}{16} + C \]
where \( C \) is the constant of integration.

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