We will solve the integral $\int_1^e x^5 \ln(x) \, dx$ using integration by parts.

Step 1: Choose $u$ and $dv$

The rule for integration by parts is:

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

We need to choose $u$ and $dv$ such that differentiating $u$ simplifies the expression.

Let:

• $u = \ln(x)$ (since the derivative of $\ln(x)$ is simpler)
• $dv = x^5 \, dx$

Step 2: Find $du$ and $v$

Now, differentiate $u$ and integrate $dv$:

• $du = \frac{1}{x} \, dx$
• $v = \frac{x^6}{6}$ (since $\int x^5 \, dx = \frac{x^6}{6}$)

Step 3: Apply the integration by parts formula

Using the formula $\int u \, dv = uv - \int v \, du$, we have:

$\int_1^e x^5 \ln(x) \, dx = \left[ \frac{x^6 \ln(x)}{6} \right]_1^e - \int_1^e \frac{x^6}{6} \cdot \frac{1}{x} \, dx$

Step 4: Simplify and evaluate the integral

First, simplify the remaining integral:

$\int_1^e \frac{x^6}{6} \cdot \frac{1}{x} \, dx = \int_1^e \frac{x^5}{6} \, dx = \frac{1}{6} \int_1^e x^5 \, dx$

Now, compute the integral $\int x^5 \, dx$:

$\int x^5 \, dx = \frac{x^6}{6}$

So, we get:

$\frac{1}{6} \int_1^e x^5 \, dx = \frac{1}{6} \left[ \frac{x^6}{6} \right]_1^e = \frac{1}{6} \left( \frac{e^6}{6} - \frac{1^6}{6} \right) = \frac{1}{6} \cdot \frac{e^6 - 1}{6} = \frac{e^6 - 1}{36}$

Step 5: Evaluate the boundary terms

Next, evaluate the boundary terms:

$\left[ \frac{x^6 \ln(x)}{6} \right]_1^e = \frac{e^6 \ln(e)}{6} - \frac{1^6 \ln(1)}{6}$

Since $\ln(e) = 1$ and $\ln(1) = 0$, this becomes:

$\frac{e^6 \cdot 1}{6} - 0 = \frac{e^6}{6}$

Step 6: Combine the results

Now, combine all the terms:

$\int_1^e x^5 \ln(x) \, dx = \frac{e^6}{6} - \frac{e^6 - 1}{36}$

To simplify, let's find a common denominator:

$\frac{e^6}{6} = \frac{6e^6}{36}$

Thus, we have:

$\int_1^e x^5 \ln(x) \, dx = \frac{6e^6}{36} - \frac{e^6 - 1}{36} = \frac{6e^6 - (e^6 - 1)}{36} = \frac{6e^6 - e^6 + 1}{36} = \frac{5e^6 + 1}{36}$

Final Answer:

$\int_1^e x^5 \ln(x) \, dx = \frac{5e^6 + 1}{36}$

Would you like more details on any of the steps? Here are 5 related questions for further practice:

1. How would the integral change if the limits were from 1 to 2 instead of 1 to $e$?
2. What is the result of differentiating $\int x^5 \ln(x) \, dx$?
3. How would the solution change if the integrand was $x^4 \ln(x)$ instead?
4. Can this integral be solved using substitution instead of integration by parts?
5. How does changing the power of $x$ in the integrand affect the solution?

Tip: When solving integration by parts, always try to pick $u$ as a function that becomes simpler when differentiated!