Answer is: option2
\(\frac{4e^5 + 1}{25}\)
Solution:
We are tasked with solving the definite integral:
\[ \int_1^e x^4 \ln(x) \, dx \]
Step 1: Differentiate and Integrate
We choose \( u = \ln(x) \) and \( dv = x^4 \, dx \).
Differentiate \( u \) and integrate \( dv \):
\[ u = \ln(x) \quad \Rightarrow \quad du = \frac{1}{x} \, dx \]
\[ dv = x^4 \, dx \quad \Rightarrow \quad v = \frac{x^5}{5} \]
Step 2: Apply the Integration by Parts Formula
Now, apply the integration by parts formula:
\[ \int_1^e x^4 \ln(x) \, dx = \left[ \ln(x) \cdot \frac{x^5}{5} \right]_1^e - \int_1^e \frac{x^5}{5} \cdot \frac{1}{x} \, dx \]
Step 3: Simplify the Expression
Simplify the remaining integral:
\[ \int_1^e \frac{x^5}{5} \cdot \frac{1}{x} \, dx = \frac{1}{5} \int_1^e x^4 \, dx \]
Now, compute \( \int_1^e x^4 \, dx \):
\[ \int_1^e x^4 \, dx = \left[ \frac{x^5}{5} \right]_1^e = \frac{e^5}{5} - \frac{1}{5} \]
So, we have:
\[ \int_1^e \frac{x^5}{5} \cdot \frac{1}{x} \, dx = \frac{1}{5} \left( \frac{e^5}{5} - \frac{1}{5} \right) = \frac{e^5 - 1}{25} \]
Step 4: Combine the Results
Now, combine the results:
\[ \int_1^e x^4 \ln(x) \, dx = \left[ \frac{e^5}{5} \ln(e) - \frac{1}{5} \ln(1) \right] - \frac{e^5 - 1}{25} \]
Since \( \ln(e) = 1 \) and \( \ln(1) = 0 \), this simplifies to:
\[ \int_1^e x^4 \ln(x) \, dx = \frac{e^5}{5} - \frac{e^5 - 1}{25} \]
Step 5: Simplify the Final Expression
Now, simplify the expression:
\[ \frac{e^5}{5} - \frac{e^5 - 1}{25} = \frac{5e^5}{25} - \frac{e^5 - 1}{25} = \frac{5e^5 - e^5 + 1}{25} = \frac{4e^5 + 1}{25} \]
Final Answer
\[ \frac{4e^5 + 1}{25} \]