Evaluate the following integral using integration by parts.

Integral from 1 to e Superscript 5 x Superscript 5 Baseline ln x dx

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!

Evaluate the following integral using integration by parts. Integral from 1 to e x^5 ln(x) dx

This problem involves evaluating the integral of x^5 ln(x) from 1 to e using the integration by parts technique. The solution covers essential calculus concepts such as logarithmic functions, polynomial integration, and the use of the integration by parts formula. Suitable for undergraduate students studying calculus.

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