Grok vs Qwen for Math

Step-by-Step Solution to ∫ x² ln(x) dx

To solve this indefinite integral, use integration by parts. The formula is:

∫ u dv = u v − ∫ v du

1. Choose u = ln(x) (this gets simpler upon differentiation)
   Then du = (1/x) dx.

2. Choose dv = x² dx
   Then v = ∫ x² dx = (x³/3).

3. Apply the integration by parts formula:
   ∫ x² ln(x) dx = [ln(x) · (x³/3)] − ∫ (x³/3) · (1/x) dx
   = (x³/3) ln(x) − (1/3) ∫ x² dx.

4. Solve the remaining integral:
   ∫ x² dx = x³/3,
   so − (1/3) · (x³/3) = −x³/9.

5. Add the constant of integration:
   ∫ x² ln(x) dx = (x³/3) ln(x) − x³/9 + C.

Verification by Differentiation

Differentiate the result y = (x³/3) ln(x) − x³/9 + C to confirm it matches the integrand x² ln(x).

• Use the product rule on the first term: d/dx [(x³/3) ln(x)]
  = [(x³/3)'] ln(x) + (x³/3) [ln(x)]'
  = (x²) ln(x) + (x³/3)(1/x)
  = x² ln(x) + (x²/3).

• Differentiate the second term: d/dx [−x³/9] = −(3x²/9) = −x²/3.

• The constant C differentiates to 0.

• Combine: x² ln(x) + x²/3 − x²/3 = x² ln(x).

The derivative matches the original integrand, so the solution is verified.

Here is the step-by-step solution to the integral $\int x^2 \ln(x) , dx$, followed by the verification.

Part 1: Solving the Integral

To solve this integral, we use Integration by Parts. The formula is: $$\int u \, dv = uv - \int v \, du$$

Step 1: Choose $u$ and $dv$ Using the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential), we prioritize the Logarithmic function for $u$.

• Let $u = \ln(x)$
• Let $dv = x^2 , dx$

Step 2: Differentiate $u$ and Integrate $dv$

• Differentiate $u$: $$du = \frac{1}{x} \, dx$$
• Integrate $dv$: $$v = \int x^2 \, dx = \frac{x^3}{3}$$

Step 3: Apply the Integration by Parts formula Substitute $u$, $v$, $du$, and $dv$ into the formula: $$\int x^2 \ln(x) \, dx = \left( \ln(x) \cdot \frac{x^3}{3} \right) - \int \left( \frac{x^3}{3} \cdot \frac{1}{x} \right) \, dx$$

Step 4: Simplify and solve the remaining integral First, simplify the expression inside the integral: $$\frac{x^3}{3} \cdot \frac{1}{x} = \frac{x^2}{3}$$

Now, update the equation: $$= \frac{x^3}{3} \ln(x) - \int \frac{x^2}{3} \, dx$$

Pull out the constant $\frac{1}{3}$: $$= \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2 \, dx$$

Integrate $x^2$: $$= \frac{x^3}{3} \ln(x) - \frac{1}{3} \left( \frac{x^3}{3} \right) + C$$

Step 5: Final Result $$= \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$$

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Part 2: Verification by Differentiation

To verify the answer, we differentiate the result with respect to $x$ and check if we get back the original integrand $x^2 \ln(x)$.

Function to differentiate: $$y = \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$$

Step 1: Differentiate the first term using the Product Rule Let $f(x) = \frac{x^3}{3}$ and $g(x) = \ln(x)$. The product rule states: $(fg)' = f'g + fg'$.

• $f'(x) = \frac{3x^2}{3} = x^2$
• $g'(x) = \frac{1}{x}$

$$\frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) \right] = (x^2)(\ln(x)) + \left( \frac{x^3}{3} \right) \left( \frac{1}{x} \right)$$ $$= x^2 \ln(x) + \frac{x^2}{3}$$

Step 2: Differentiate the second term $$\frac{d}{dx} \left[ -\frac{x^3}{9} \right] = -\frac{3x^2}{9} = -\frac{x^2}{3}$$

Step 3: Differentiate the constant $$\frac{d}{dx} [C] = 0$$

Step 4: Combine the results $$\frac{dy}{dx} = \left( x^2 \ln(x) + \frac{x^2}{3} \right) - \frac{x^2}{3} + 0$$

The terms $+\frac{x^2}{3}$ and $-\frac{x^2}{3}$ cancel out: $$\frac{dy}{dx} = x^2 \ln(x)$$

Conclusion: The derivative matches the original integrand, confirming that the solution is correct.

Final Answer: $$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$$