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We are tasked with solving the following integral:

\[
I = \int_0^1 x \ln(x + 1) \, dx
\]

### Approach:

We will use **integration by parts**, which is generally written as:

\[
\int u \, dv = uv - \int v \, du
\]

We need to choose \( u \) and \( dv \) such that the integration simplifies. Let's choose:
- \( u = \ln(x + 1) \), then \( du = \frac{1}{x + 1} \, dx \)
- \( dv = x \, dx \), then \( v = \frac{x^2}{2} \)

Now, applying the integration by parts formula:

\[
I = \left[ \frac{x^2}{2} \ln(x+1) \right]_0^1 - \int_0^1 \frac{x^2}{2} \cdot \frac{1}{x+1} \, dx
\]

### Step 1: Evaluate the boundary term

First, we evaluate the boundary term \( \frac{x^2}{2} \ln(x+1) \) from \( 0 \) to \( 1 \):

At \( x = 1 \):
\[
\frac{1^2}{2} \ln(1+1) = \frac{1}{2} \ln(2)
\]

At \( x = 0 \):
\[
\frac{0^2}{2} \ln(0+1) = 0
\]

So, the boundary term is \( \frac{1}{2} \ln(2) \).

### Step 2: Solve the remaining integral

We now deal with the remaining integral:

\[
J = \int_0^1 \frac{x^2}{2(x+1)} \, dx
\]

This can be simplified and evaluated using partial fraction decomposition or by recognizing a standard form. We can perform that step next.The remaining integral evaluates to:

\[
J = \int_0^1 \frac{x^2}{2(x+1)} \, dx = \frac{\ln(2)}{2} - \frac{1}{4}
\]

### Final Answer:

Now, combining everything, the integral becomes:

\[
I = \frac{1}{2} \ln(2) - \left( \frac{\ln(2)}{2} - \frac{1}{4} \right)
\]

Simplifying the expression:

\[
I = \frac{1}{4}
\]

Thus, the value of the integral is:

\[
\int_0^1 x \ln(x + 1) \, dx = \frac{1}{4}
\]

Would you like further details on any step? Here are some related questions to deepen your understanding:

1. How does integration by parts work, and why was it chosen here?
2. Can we solve this integral using another method, such as substitution?
3. What is partial fraction decomposition, and how can it simplify integrals?
4. How do logarithmic properties assist in integrating functions like \( \ln(x + 1) \)?
5. What are the general steps to approach integrals involving products of \( x \) and logarithmic functions?

**Tip:** When solving integrals with logarithmic functions, integration by parts is often a powerful tool, as it reduces the complexity of the logarithmic terms.

Learn how to solve the integral \( \int_0^1 x \ln(x + 1) \, dx \) using integration by parts. This problem involves key calculus concepts, such as integration by parts and logarithmic properties. Detailed step-by-step solution provided. Suitable for undergraduate-level students.

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