Kimi vs Qwen for Math

Here is the step-by-step solution using integration by parts.

Step 1: Choose $u$ and $dv$

For $\int x^2 \ln(x) , dx$, we use the integration by parts formula: $$\int u \, dv = uv - \int v \, du$$

We choose:

• $u = \ln(x)$  (since it simplifies when differentiated)
• $dv = x^2 , dx$  (since it's easy to integrate)

This gives us:

• $du = \frac{1}{x} , dx$
• $v = \frac{x^3}{3}$

Step 2: Apply the formula

$$\int x^2 \ln(x) \, dx = uv - \int v \, du$$ $$= \ln(x) \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx$$

Step 3: Simplify the remaining integral

$$= \frac{x^3 \ln(x)}{3} - \int \frac{x^2}{3} \, dx$$ $$= \frac{x^3 \ln(x)}{3} - \frac{1}{3} \int x^2 \, dx$$ $$= \frac{x^3 \ln(x)}{3} - \frac{1}{3} \cdot \frac{x^3}{3} + C$$

Step 4: Final Answer

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

(Or factored form: $\frac{x^3}{9}(3\ln(x) - 1) + C$)

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Verification by Differentiation

Let's differentiate $F(x) = \frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C$:

Term 1: $\frac{d}{dx}\left[\frac{x^3 \ln(x)}{3}\right]$ Using the product rule on $x^3 \ln(x)$: $$= \frac{1}{3}\left[3x^2 \cdot \ln(x) + x^3 \cdot \frac{1}{x}\right] = \frac{1}{3}\left[3x^2 \ln(x) + x^2\right] = x^2 \ln(x) + \frac{x^2}{3}$$

Term 2: $\frac{d}{dx}\left[-\frac{x^3}{9}\right] = -\frac{3x^2}{9} = -\frac{x^2}{3}$

Term 3: $\frac{d}{dx}[C] = 0$

Adding together: $$\left(x^2 \ln(x) + \frac{x^2}{3}\right) + \left(-\frac{x^2}{3}\right) + 0 = x^2 \ln(x)$$

Since $\frac{d}{dx}\left[\frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C\right] = x^2 \ln(x)$, the solution is verified.

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

Part 1: Solving the Integral

Since the integrand is the product of an algebraic function ($x^2$) and a logarithmic function ($\ln(x)$), the most effective method is Integration by Parts.

The formula for Integration by Parts is: $$\int u \, dv = uv - \int v \, du$$

Step 1: Choose $u$ and $dv$ We use the LIATE rule (Logarithmic, Inverse Trig, Algebraic, Trig, Exponential) to select $u$. Since Logarithmic functions come before Algebraic functions, we set:

• $u = \ln(x)$
• $dv = x^2 , dx$

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

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

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

Step 4: Simplify and Integrate Simplify the expression inside the integral: $$= \frac{x^3}{3}\ln(x) - \int \frac{x^2}{3} \, dx$$

Factor out the constant $\frac{1}{3}$ and solve the remaining integral using the power rule: $$= \frac{x^3}{3}\ln(x) - \frac{1}{3} \int x^2 \, dx$$ $$= \frac{x^3}{3}\ln(x) - \frac{1}{3} \left(\frac{x^3}{3}\right) + C$$

Final Simplified Answer: $$\frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C$$

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Part 2: Verification by Differentiation

To verify the result, we differentiate the solution with respect to $x$. We should arrive back at the original integrand, $x^2 \ln(x)$.

$$\frac{d}{dx} \left[ \frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C \right]$$

Step 1: Differentiate the First Term (Product Rule) Apply the product rule, $(fg)' = f'g + fg'$, to the first term:

• Let $f = \frac{x^3}{3}$, then $f' = x^2$.
• Let $g = \ln(x)$, then $g' = \frac{1}{x}$.

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

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

Step 3: Combine and Simplify Combine the results from Steps 1 and 2 (the derivative of the constant $C$ is 0): $$\left( x^2 \ln(x) + \frac{x^2}{3} \right) - \frac{x^2}{3}$$

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

Conclusion: The derivative of our result is exactly the original integrand. The solution is correct.