Gemini vs Qwen for Math

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

Part 1: Evaluating the Integral

We want to solve the integral: ∫(x² · ln(x)) dx

Since we are integrating the product of two different types of functions (algebraic and logarithmic), we should use Integration by Parts. The formula for integration by parts is: ∫ u dv = u·v - ∫ v du

Step 1: Choose u and dv. To decide which function should be $u$, we can use the LIATE acronym (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). Logarithmic comes before Algebraic, so:

• Let u = ln(x)
• Let dv = x² dx

Step 2: Differentiate u and integrate dv.

• Differentiate $u$ to find $du$: du = $\frac{1}{x}$ dx
• Integrate $dv$ to find $v$: v = ∫ x² dx = $\frac{x^3}{3}$

Step 3: Apply the integration by parts formula. ∫(x² · ln(x)) dx = (ln(x)) · ($\frac{x^3}{3}$) - ∫ ($\frac{x^3}{3}$) · ($\frac{1}{x}$) dx

Step 4: Simplify and solve the new integral. = $\frac{x^3}{3}$ ln(x) - ∫ $\frac{x^2}{3}$ dx

Pull the constant ($\frac{1}{3}$) out of the integral: = $\frac{x^3}{3}$ ln(x) - $\frac{1}{3}$ ∫ x² dx

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

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

(Optional: You can also factor out $\frac{x^3}{9}$ to write it as $\frac{x^3}{9}$(3 ln(x) - 1) + C)

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Part 2: Verifying the Answer

To verify, we will take the derivative of our result with respect to $x$. If our result is correct, the derivative will equal the original integrand, $x^2 \ln(x)$.

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

Step 1: Apply the derivative rules.

• For the first term ($\frac{x^3}{3}$ ln(x)), we must use the Product Rule: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
• For the second term (-$\frac{x^3}{9}$), we use the basic power rule.
• For the constant (C), the derivative is 0.

Step 2: Differentiate the first term using the Product Rule.

• Let f(x) = $\frac{x^3}{3}$, so f'(x) = $x^2$
• Let g(x) = ln(x), so g'(x) = $\frac{1}{x}$
• Apply the rule: ($x^2$)(ln(x)) + ($\frac{x^3}{3}$)($\frac{1}{x}$) = $x^2$ ln(x) + $\frac{x^2}{3}$

Step 3: Differentiate the second term.

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

Step 4: Combine everything. Derivative = [ $x^2$ ln(x) + $\frac{x^2}{3}$ ] - $\frac{x^2}{3}$ + 0

Notice that the $+\frac{x^2}{3}$ and $-\frac{x^2}{3}$ cancel each other out.

Derivative = x² · ln(x)

Because taking the derivative mathematically brings us perfectly back to the original function inside the integral, our answer is verified as exactly correct!

To solve the integral $\int x^2 \ln(x) , dx$, we will use the method of Integration by Parts.

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

Part 1: Solve the Integral

Step 1: Choose $u$ and $dv$. We follow the LIATE rule (Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential) to choose $u$. Since Logarithmic comes before Algebraic, we set:

• $u = \ln(x)$
• $dv = x^2 , dx$

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

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

Step 4: Simplify the expression. $$\frac{x^3}{3} \ln(x) - \int \frac{x^2}{3} \, dx$$

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

Step 5: Solve the remaining integral. We know that $\int x^2 , dx = \frac{x^3}{3}$. $$\frac{x^3}{3} \ln(x) - \frac{1}{3} \left( \frac{x^3}{3} \right) + C$$

Step 6: Final Simplification. $$\frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$$

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Part 2: Verification by Differentiation

To verify the result, we differentiate the answer with respect to $x$. The result should match the original integrand ($x^2 \ln(x)$).

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

Step 1: Apply the Sum/Difference Rule. $$\frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) \right] - \frac{d}{dx} \left[ \frac{x^3}{9} \right] + \frac{d}{dx} [C]$$

Step 2: Differentiate the first term using the Product Rule. Let $f = \frac{x^3}{3}$ and $g = \ln(x)$. Then $f' = x^2$ and $g' = \frac{1}{x}$. Using $(fg)' = f'g + fg'$: $$\frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) \right] = (x^2)(\ln(x)) + \left( \frac{x^3}{3} \right) \left( \frac{1}{x} \right)$$ $$= x^2 \ln(x) + \frac{x^2}{3}$$

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

Step 4: Differentiate the constant. $$\frac{d}{dx} [C] = 0$$

Step 5: Combine all parts. $$\frac{dy}{dx} = \left( x^2 \ln(x) + \frac{x^2}{3} \right) - \frac{x^2}{3} + 0$$

Step 6: Simplify. The terms $+\frac{x^2}{3}$ and $-\frac{x^2}{3}$ cancel out: $$\frac{dy}{dx} = x^2 \ln(x)$$

Since the derivative matches the original integrand, the solution is correct.

Final Answer:

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