To solve the integral $\int x \ln(3 + x) \, dx$, we'll first make a substitution, and then apply integration by parts.

Step 1: Substitution

Let’s set: $u = 3 + x$ This implies: $du = dx$ Also, $x = u - 3$, so the integral becomes: $\int (u - 3) \ln(u) \, du$ We can now split the integral into two parts: $\int u \ln(u) \, du - 3 \int \ln(u) \, du$

Step 2: Integration by Parts for Each Integral

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

To solve $\int u \ln(u) \, du$, we use integration by parts:

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

Now, applying the integration by parts formula $\int v \, dw = v w - \int w \, dv$, we get: $\int u \ln(u) \, du = \frac{u^2}{2} \ln(u) - \int \frac{u^2}{2} \cdot \frac{1}{u} \, du$ $= \frac{u^2}{2} \ln(u) - \int \frac{u}{2} \, du$ $= \frac{u^2}{2} \ln(u) - \frac{u^2}{4}$

Second Integral: $-3 \int \ln(u) \, du$

We use integration by parts again:

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

Thus: $\int \ln(u) \, du = u \ln(u) - u$ Multiplying by $-3$, we get:

= -3u \ln(u) + 3u$$ ### Step 3: Combine the Results Now, adding both parts, the integral becomes: $$\frac{u^2}{2} \ln(u) - \frac{u^2}{4} - 3u \ln(u) + 3u + C$$ Finally, substitute $$u = 3 + x$$ back: $$= \frac{(3 + x)^2}{2} \ln(3 + x) - \frac{(3 + x)^2}{4} - 3(3 + x) \ln(3 + x) + 3(3 + x) + C$$ This is the evaluated integral. Would you like further clarification or details on any specific part? ### Related Questions: 1. How is integration by parts derived from the product rule? 2. Can integration by parts be applied to trigonometric functions? 3. What are the applications of substitution in integral calculus? 4. How do you handle integrals involving logarithmic functions? 5. What are some common errors to avoid in integration by parts? ### Tip: Always double-check your substitution and integration by parts setup to avoid calculation errors later in the process.