To compute the indefinite integral of $x^5 \ln(x) \, dx$, we will use integration by parts. The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

Step 1: Assign $u$ and $dv$

Let: $u = \ln(x) \quad \text{(since its derivative is simpler)}$ $dv = x^5 \, dx$

Step 2: Compute $du$ and $v$

The derivative of $u$ is: $du = \frac{1}{x} \, dx$

The integral of $dv$ is: $v = \int x^5 \, dx = \frac{x^6}{6}$

Step 3: Apply the integration by parts formula

Substitute into the formula: $\int x^5 \ln(x) \, dx = uv - \int v \, du$ $= \ln(x) \cdot \frac{x^6}{6} - \int \frac{x^6}{6} \cdot \frac{1}{x} \, dx$

Step 4: Simplify the second integral

Simplify the terms: $\int x^5 \ln(x) \, dx = \frac{x^6 \ln(x)}{6} - \frac{1}{6} \int x^5 \, dx$

The second integral is: $\int x^5 \, dx = \frac{x^6}{6}$

So, substitute this result: $\int x^5 \ln(x) \, dx = \frac{x^6 \ln(x)}{6} - \frac{1}{6} \cdot \frac{x^6}{6}$

Step 5: Simplify further

$\int x^5 \ln(x) \, dx = \frac{x^6 \ln(x)}{6} - \frac{x^6}{36}$

Step 6: Combine terms and add the constant of integration

Factor out $\frac{x^6}{6}$: $\int x^5 \ln(x) \, dx = \frac{x^6}{6} \left( \ln(x) - \frac{1}{6} \right) + C$

Final Answer:

$\boxed{\int x^5 \ln(x) \, dx = \frac{x^6}{6} \left( \ln(x) - \frac{1}{6} \right) + C}$

Let me know if you want detailed steps explained further!

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Relative Questions:

1. What is the significance of using integration by parts in calculus?
2. How do you determine the assignments for $u$ and $dv$ in integration by parts?
3. Can you apply this technique to other logarithmic functions?
4. How do you handle integrals involving both exponential and logarithmic terms?
5. What are the alternative approaches to solve this integral if any exist?

Tip: When using integration by parts, always simplify $du$ and $dv$ as much as possible to reduce the complexity of the remaining integral.