Let's evaluate the integral $\int x \ln(3 + x) \, dx$ using substitution and integration by parts.

Step 1: Substitution

We begin by using substitution to simplify the integral.

Let: $u = 3 + x \quad \Rightarrow \quad du = dx$ Now, rewrite the integral in terms of $u$. Notice that $x = u - 3$, so the integral becomes: $\int x \ln(3 + x) \, dx = \int (u - 3) \ln(u) \, du$

Step 2: Integration by Parts

Next, we use the integration by parts formula: $\int v \, dw = v w - \int w \, dv$ In this case, let:

• $v = \ln(u)$, so $dv = \frac{1}{u} \, du$
• $dw = (u - 3) \, du$, so we need to integrate $dw$.

We can break $(u - 3)$ into two terms: $\int (u - 3) \ln(u) \, du = \int u \ln(u) \, du - 3 \int \ln(u) \, du$ Now we apply integration by parts to each term.

First term: $\int u \ln(u) \, du$

Let:

• $v = \ln(u)$, so $dv = \frac{1}{u} \, du$
• $dw = u \, du$, so $w = \frac{u^2}{2}$

Using the integration by parts formula: $\int u \ln(u) \, du = \frac{u^2}{2} \ln(u) - \int \frac{u^2}{2} \cdot \frac{1}{u} \, du = \frac{u^2}{2} \ln(u) - \frac{1}{2} \int u \, du$ Now, integrate $u$: $\frac{1}{2} \int u \, du = \frac{1}{2} \cdot \frac{u^2}{2} = \frac{u^2}{4}$ Thus: $\int u \ln(u) \, du = \frac{u^2}{2} \ln(u) - \frac{u^2}{4}$

Second term: $\int \ln(u) \, du$

Using integration by parts again: Let:

• $v = \ln(u)$, so $dv = \frac{1}{u} \, du$
• $dw = du$, so $w = u$

Thus: $\int \ln(u) \, du = u \ln(u) - \int u \cdot \frac{1}{u} \, du = u \ln(u) - u$

Step 3: Putting It All Together

Now substitute both results into the original integral: $\int (u - 3) \ln(u) \, du = \left( \frac{u^2}{2} \ln(u) - \frac{u^2}{4} \right) - 3 \left( u \ln(u) - u \right)$ Simplify: $= \frac{u^2}{2} \ln(u) - \frac{u^2}{4} - 3u \ln(u) + 3u$

Step 4: Substituting Back for $u$

Recall that $u = 3 + x$. Substituting back into the equation: $\int x \ln(3 + x) \, dx = \frac{(3 + x)^2}{2} \ln(3 + x) - \frac{(3 + x)^2}{4} - 3(3 + x) \ln(3 + x) + 3(3 + x)$

Step 5: Simplifying

Now, collect like terms: $= \left( \frac{(3 + x)^2}{2} \ln(3 + x) - 3(3 + x) \ln(3 + x) \right) + \left( - \frac{(3 + x)^2}{4} + 3(3 + x) \right)$ Factor terms involving $\ln(3 + x)$ and the constants.

Final Answer

After simplifying the above expression, don't forget the constant of integration, $C$, since we are solving an indefinite integral.

Thus, the final result is: $\boxed{\int x \ln(3 + x) \, dx = \frac{(3 + x)^2}{2} \ln(3 + x) - \frac{(3 + x)^2}{4} - 3(3 + x) \ln(3 + x) + 3(3 + x) + C}$

---

Would you like further details on any of the steps, or have any specific questions? Here are 5 additional questions for you:

1. How would you apply integration by parts to more complicated logarithmic integrals?
2. What are some common substitution strategies for simplifying integrals involving logarithms?
3. Could this technique be applied to other functions like $e^x$ or trigonometric functions?
4. What is the general formula for integration by parts, and how can it be memorized?
5. How do you know when to use substitution or integration by parts in a given integral?

Tip: When dealing with integrals involving products of functions like $x \ln(f(x))$, substitution can simplify the problem significantly before applying integration by parts.