The integral in question is:

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

We are asked to compute this using integration by parts. Let's proceed step by step.

Step 1: Integration by parts formula

The formula for integration by parts is:

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

In this case, we can choose:

• $u = \ln(x)$ (so that its derivative simplifies)
• $dv = x \, dx$ (this will make the integral simpler)

Step 2: Compute $du$ and $v$

• $u = \ln(x)$ implies $du = \frac{1}{x} \, dx$
• $dv = x \, dx$ implies $v = \frac{x^2}{2}$

Step 3: Apply the integration by parts formula

Now, apply the formula $\int u \, dv = uv - \int v \, du$:

$I = \left[ \ln(x) \cdot \frac{x^2}{2} \right]_1^e - \int_1^e \frac{x^2}{2} \cdot \frac{1}{x} \, dx$

This simplifies to:

$I = \left[ \frac{x^2 \ln(x)}{2} \right]_1^e - \frac{1}{2} \int_1^e x \, dx$

Step 4: Evaluate each term

1. Evaluate $\left[ \frac{x^2 \ln(x)}{2} \right]_1^e$:
   • At $x = e$, $\frac{e^2 \ln(e)}{2} = \frac{e^2}{2}$ (since $\ln(e) = 1$)
   • At $x = 1$, $\frac{1^2 \ln(1)}{2} = 0$ (since $\ln(1) = 0$)

Thus, $\left[ \frac{x^2 \ln(x)}{2} \right]_1^e = \frac{e^2}{2} - 0 = \frac{e^2}{2}$.

2. Evaluate $\int_1^e x \, dx$: $\int_1^e x \, dx = \left[ \frac{x^2}{2} \right]_1^e = \frac{e^2}{2} - \frac{1^2}{2} = \frac{e^2}{2} - \frac{1}{2}$

So, $\frac{1}{2} \int_1^e x \, dx = \frac{1}{2} \left( \frac{e^2}{2} - \frac{1}{2} \right) = \frac{e^2}{4} - \frac{1}{4}$.

Step 5: Combine the results

Now, combine the two results to find $I$:

$I = \frac{e^2}{2} - \left( \frac{e^2}{4} - \frac{1}{4} \right)$

Simplifying:

$I = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{2e^2}{4} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4}$

Thus, the value of $I$ is:

$I = \frac{e^2 + 1}{4}$

Correct answer:

The correct answer is C.

Would you like further details or clarifications on this? Here are 5 related questions:

1. How can you determine the best choice of $u$ and $dv$ in integration by parts?
2. What other methods can be used to solve integrals involving logarithmic functions?
3. How does integration by parts simplify certain types of integrals?
4. What are the applications of integrals in real-world problems?
5. Why does $\ln(1) = 0$, and how does that simplify calculations?

Tip: Integration by parts is particularly useful when one of the functions becomes simpler after differentiation, such as logarithmic or inverse trigonometric functions.