To solve the integral $\int x \ln(1 + x) \, dx$, we can apply integration by parts.

Recall the formula for integration by parts:

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

We will choose $u = \ln(1 + x)$ and $dv = x \, dx$.

Step 1: Differentiate $u$ and integrate $dv$

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

Step 2: Apply the integration by parts formula

Now substitute into the integration by parts formula:

$\int x \ln(1 + x) \, dx = \frac{x^2}{2} \ln(1 + x) - \int \frac{x^2}{2(1 + x)} \, dx$

Step 3: Simplify the remaining integral

The remaining integral is:

$\int \frac{x^2}{2(1 + x)} \, dx$

We can simplify $\frac{x^2}{1 + x}$ as follows:

$\frac{x^2}{1 + x} = x - 1 + \frac{1}{1 + x}$

Thus, the integral becomes:

$\int \frac{x^2}{2(1 + x)} \, dx = \frac{1}{2} \int (x - 1 + \frac{1}{1 + x}) \, dx$

This integral can now be separated:

$\frac{1}{2} \int x \, dx - \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{1}{1 + x} \, dx$

Step 4: Solve each part

• $\int x \, dx = \frac{x^2}{2}$
• $\int 1 \, dx = x$
• $\int \frac{1}{1 + x} \, dx = \ln(1 + x)$

Step 5: Combine the results

Now, substitute the integrals back:

$\frac{1}{2} \left( \frac{x^2}{2} - x + \ln(1 + x) \right)$

Step 6: Final answer

Substitute this result into the original equation:

$\int x \ln(1 + x) \, dx = \frac{x^2}{2} \ln(1 + x) - \frac{1}{2} \left( \frac{x^2}{2} - x + \ln(1 + x) \right) + C$

Simplifying:

$\int x \ln(1 + x) \, dx = \frac{x^2}{2} \ln(1 + x) - \frac{x^2}{4} + \frac{x}{2} - \frac{\ln(1 + x)}{2} + C$

This is the final result.

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Would you like further clarification or details on any step?

Here are 5 related questions to extend this topic:

1. How does integration by parts work for more complex functions?
2. What is the intuition behind choosing $u$ and $dv$ in integration by parts?
3. How would you solve the integral $\int \ln(1 + x) \, dx$?
4. What is the method for simplifying integrals involving logarithmic functions?
5. Can substitution be combined with integration by parts in certain integrals?

Tip: When applying integration by parts, it's often helpful to choose $u$ as the term that simplifies when differentiated.