Gemini vs Kimi for Math

Here is the step-by-step solution to evaluate the integral, followed by the verification.

Part 1: Evaluating the Integral

To solve the integral $\int x^2 \ln(x) , dx$, we will use Integration by Parts. The formula for integration by parts is: $\int u , dv = u \cdot v - \int v , du$

Step 1: Choose $u$ and $dv$. We use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to determine our $u$. Since we have a logarithmic function ($\ln(x)$) and an algebraic function ($x^2$), we set:

• $u = \ln(x)$
• $dv = x^2 , dx$

Step 2: Find $du$ and $v$.

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

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

Step 4: Simplify and solve the remaining integral. First, clean up the expression: $= \frac{x^3}{3} \ln(x) - \int \frac{x^2}{3} , dx$

Now, pull the constant ($\frac{1}{3}$) out of the integral and integrate $x^2$: $= \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2 , dx$ $= \frac{x^3}{3} \ln(x) - \frac{1}{3} \left(\frac{x^3}{3}\right) + C$

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

(Note: This can also be factored and written as $\frac{x^3}{9}(3\ln(x) - 1) + C$)

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Part 2: Verification by Differentiation

To verify our answer, we must take the derivative of our result. If correct, it should evaluate back to the original integrand, $x^2 \ln(x)$.

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

Step 1: Differentiate the first term using the Product Rule. The Product Rule is $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$. Let $f(x) = \frac{x^3}{3}$ and $g(x) = \ln(x)$.

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

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

Step 2: Differentiate the second and third terms.

• $\frac{d}{dx} \left[ -\frac{x^3}{9} \right] = -\frac{3x^2}{9} = -\frac{x^2}{3}$
• $\frac{d}{dx} [C] = 0$

Step 3: Combine all the derivatives. $\frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C \right] = \left( x^2 \ln(x) + \frac{x^2}{3} \right) - \frac{x^2}{3} + 0$

Notice that $+ \frac{x^2}{3}$ and $- \frac{x^2}{3}$ cancel each other out perfectly: $= x^2 \ln(x)$

Because taking the derivative yields our original integrand, our integration is verified to be correct.

Step 1: Identify the integration technique

To solve $\int x^2\ln(x),dx$, we use integration by parts since we have a product of an algebraic function ($x^2$) and a logarithmic function ($\ln(x)$).

The formula is: $$\int u\,dv = uv - \int v\,du$$

Step 2: Choose $u$ and $dv$

Following the LIATE rule (Logarithmic before Algebraic), we choose:

• $u = \ln(x)$
• $dv = x^2,dx$

Step 3: Calculate $du$ and $v$

Differentiate $u$: $$du = \frac{1}{x}\,dx$$

Integrate $dv$: $$v = \int x^2\,dx = \frac{x^3}{3}$$

Step 4: Apply the integration by parts formula

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

Step 5: Simplify and solve 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 6: Final answer

$$= \frac{x^3\ln(x)}{3} - \frac{x^3}{9} + C$$

Or factored: $$\boxed{\int x^2\ln(x)\,dx = \frac{x^3}{9}\left(3\ln(x) - 1\right) + C}$$

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Verification by Differentiation

Let $F(x) = \frac{x^3\ln(x)}{3} - \frac{x^3}{9} + C$. We differentiate using the product rule on the first term:

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

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

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

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

The derivative matches the original integrand, confirming our solution is correct.