We are tasked with evaluating the integral:

$I = \int_1^e 6x^4 \ln(x) \, dx$

Step 1: Choose parts for integration by parts

We can apply the integration by parts formula:

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

We choose:

• $u = \ln(x)$, so $du = \frac{1}{x} \, dx$
• $dv = 6x^4 \, dx$, so $v = \frac{6x^5}{5}$

Step 2: Apply the formula

Now, applying the integration by parts formula:

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

Simplifying the second integral:

$I = \left[ \frac{6x^5}{5} \ln(x) \right]_1^e - \frac{6}{5} \int_1^e x^4 \, dx$

Step 3: Evaluate the boundary terms

First, let's evaluate the boundary term:

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

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

$\frac{6e^5}{5} \times 1 - 0 = \frac{6e^5}{5}$

Step 4: Evaluate the remaining integral

Next, we compute the remaining integral:

$\frac{6}{5} \int_1^e x^4 \, dx$

The integral of $x^4$ is:

$\int x^4 \, dx = \frac{x^5}{5}$

Thus:

$\frac{6}{5} \int_1^e x^4 \, dx = \frac{6}{5} \left[ \frac{x^5}{5} \right]_1^e = \frac{6}{5} \left( \frac{e^5}{5} - \frac{1^5}{5} \right)$

Simplifying:

$\frac{6}{5} \times \frac{e^5 - 1}{5} = \frac{6(e^5 - 1)}{25}$

Step 5: Final expression

Now we can combine both terms:

$I = \frac{6e^5}{5} - \frac{6(e^5 - 1)}{25}$

To combine these fractions, first get a common denominator:

$I = \frac{30e^5}{25} - \frac{6(e^5 - 1)}{25}$

Now simplify:

$I = \frac{30e^5 - 6(e^5 - 1)}{25}$

Distribute the 6:

$I = \frac{30e^5 - 6e^5 + 6}{25}$

Combine like terms:

$I = \frac{24e^5 + 6}{25}$

Thus, the value of the integral is:

$I = \frac{6(4e^5 + 1)}{25}$

Final Answer:

$I = \frac{6(4e^5 + 1)}{25}$

Would you like a more detailed explanation of any of the steps or concepts?

Here are 5 related questions you could explore:

1. How do you handle definite integrals when the integrand involves both powers of $x$ and logarithmic functions?
2. What happens if you use a different choice for $u$ and $dv$ in integration by parts?
3. How would the integral change if the limits were different, say from 0 to 1 instead of 1 to $e$?
4. How do you apply integration by parts to integrals involving exponential and logarithmic functions?
5. Can integration by parts be used to evaluate integrals involving trigonometric functions?

Tip: When choosing $u$ and $dv$ for integration by parts, it’s often helpful to pick $u$ as the function that simplifies when differentiated (like $\ln(x)$, which becomes $\frac{1}{x}$ upon differentiation).