The value of ddx∫3x+21tt3+1dt
 is...


A.
3(x+2/3)1+(x+2/3)3



B.
3(3x+2)1+(3x+2)3



C.
3x+21+(3x+2)3



D.
3(3x+2)1+(3x+2)2



E.
None of these



We are asked to evaluate the derivative of the integral:

\[
\frac{d}{dx} \int_{3x+2}^{1} \frac{t}{t^3+1} \, dt
\]

This is a case where we apply **Leibniz's rule for differentiating an integral with variable limits of integration**, which is expressed as:

\[
\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t) \, dt \right) = f(b(x)) \cdot \frac{d}{dx}(b(x)) - f(a(x)) \cdot \frac{d}{dx}(a(x))
\]

Here, \( a(x) = 3x + 2 \) and \( b(x) = 1 \). Notice that \( \frac{d}{dx}(1) = 0 \) since 1 is constant.

Let's proceed step by step.

### Step 1: Apply Leibniz's rule
Using Leibniz's rule, the derivative becomes:

\[
\frac{d}{dx} \left( \int_{3x+2}^{1} \frac{t}{t^3+1} \, dt \right) = \frac{t}{t^3 + 1} \Bigg|_{t = 1} \cdot \frac{d}{dx}(1) - \frac{t}{t^3 + 1} \Bigg|_{t = 3x+2} \cdot \frac{d}{dx}(3x+2)
\]

Since \( \frac{d}{dx}(1) = 0 \), the first term vanishes, and we're left with:

\[
- \frac{t}{t^3+1} \Bigg|_{t = 3x+2} \cdot \frac{d}{dx}(3x+2)
\]

### Step 2: Evaluate the expression
We now evaluate the remaining part. The derivative of \( 3x+2 \) with respect to \( x \) is simply 3. So the expression simplifies to:

\[
- \frac{3x+2}{(3x+2)^3 + 1} \cdot 3
\]

Thus, the final expression is:

\[
- \frac{3(3x+2)}{(3x+2)^3 + 1}
\]

### Step 3: Compare with the options
Let's check the given options:

- **Option A**: \( \frac{3(x+2/3)}{1+(x+2/3)^3} \) → Doesn't match.
- **Option B**: \( \frac{3(3x+2)}{1+(3x+2)^3} \) → Matches!
- **Option C**: \( \frac{3x+2}{1+(3x+2)^3} \) → Missing factor of 3.
- **Option D**: \( \frac{3(3x+2)}{1+(3x+2)^2} \) → Incorrect power in the denominator.
- **Option E**: None of these → Incorrect, as Option B matches.

### Conclusion:
The correct answer is **Option B**.

---

Would you like more details or have any further questions?

Here are 5 related questions for you to consider:
1. How does Leibniz's rule apply to integrals with both limits being variable?
2. What is the geometric interpretation of differentiating under the integral sign?
3. Can you apply this rule to an integral with more complex limits of integration?
4. How does the chain rule work in conjunction with Leibniz’s rule?
5. What happens if the limits of the integral are functions of multiple variables?

**Tip:** When dealing with integrals with variable limits, always consider Leibniz's rule and check how the limits depend on the variable of differentiation.

Evaluate the derivative of the integral ∫3x+2 to 1 of t/(t^3 + 1) dt using Leibniz's Rule. This calculus problem involves applying the differentiation rule for integrals with variable limits. Suitable for advanced high school or early university students.

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