To solve the integral $\int x^5 \ln(x) \, dx$, we will use the method of integration by parts.

Integration by Parts Formula:

$\int u \, dv = uv - \int v \, du$ Here, we need to choose $u$ and $dv$ from the integrand $x^5 \ln(x) \, dx$.

Step 1: Choose $u$ and $dv$

Let: $u = \ln(x) \quad \text{and} \quad dv = x^5 \, dx$

Step 2: Compute $du$ and $v$

Differentiate $u$ to find $du$: $du = \frac{1}{x} \, dx$ Integrate $dv$ to find $v$: $v = \frac{x^6}{6}$

Step 3: Apply the Integration by Parts Formula

Now, substitute into the integration by parts formula: $\int x^5 \ln(x) \, dx = \frac{x^6}{6} \ln(x) - \int \left(\frac{x^6}{6} \cdot \frac{1}{x}\right) dx$ Simplify the integrand of the remaining integral: $\int x^5 \ln(x) \, dx = \frac{x^6}{6} \ln(x) - \frac{1}{6} \int x^5 \, dx$

Step 4: Integrate the Remaining Integral

Now, integrate $\int x^5 \, dx$: $\int x^5 \, dx = \frac{x^6}{6}$ So, the integral becomes: $\int x^5 \ln(x) \, dx = \frac{x^6}{6} \ln(x) - \frac{1}{6} \cdot \frac{x^6}{6} + C$ $\int x^5 \ln(x) \, dx = \frac{x^6}{6} \ln(x) - \frac{x^6}{36} + C$

Final Answer

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

Do you want any further details or have any questions?

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Here are 5 questions to expand on this concept:

1. How does the choice of $u$ and $dv$ affect the integration by parts process?
2. Can you solve $\int x^2 \ln(x) \, dx$ using a similar method?
3. What happens if you integrate by parts multiple times?
4. How would you handle the integral $\int x \ln(x) \, dx$?
5. How can integration by parts be used in solving differential equations?

Tip: When choosing $u$ in integration by parts, it's often useful to pick the function that simplifies when differentiated, such as logarithmic or inverse trigonometric functions.