Gemini vs Grok for Math

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

Part 1: Solving the Integral

To solve the integral $\int x^2 \ln(x) , dx$, we will use a technique called integration by parts. The formula for integration by parts is:

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

Step 1: Choose $u$ and $dv$ A good rule of thumb for choosing $u$ is the acronym LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). Since we have a Logarithmic function ($\ln(x)$) and an Algebraic function ($x^2$), we choose the logarithmic function as $u$.

Let:

• $u = \ln(x)$
• $dv = x^2 , dx$

Step 2: Find $du$ and $v$ We differentiate $u$ to find $du$, and we integrate $dv$ to find $v$:

• $du = \frac{1}{x} , dx$
• $v = \int x^2 , dx = \frac{x^3}{3}$

Step 3: Apply the integration by parts formula Substitute our chosen values 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 solve the remaining integral Rewrite the first term and simplify the expression inside the new integral: $$= \frac{x^3}{3} \ln(x) - \int \frac{x^2}{3} \, dx$$

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

Now, integrate $x^2$: $$= \frac{x^3}{3} \ln(x) - \frac{1}{3} \left( \frac{x^3}{3} \right)$$

Step 5: Write the final answer Multiply the fractions and add the constant of integration, $C$: $$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$$

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Part 2: Verifying by Differentiation

To verify our answer, we will take the derivative of our result. If our integration is correct, taking the derivative should get us back to the original function: $x^2 \ln(x)$.

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

Step 1: Use the Product Rule for the first term For the expression $\frac{x^3}{3} \ln(x)$, we have the product of two functions, so we apply the product rule: $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$.

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 product 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 remaining terms Next, we differentiate the second term, $-\frac{x^3}{9}$, and the constant, $C$:

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

Step 3: Combine everything Put all the differentiated parts back together: $$\frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C \right] = x^2 \ln(x) + \frac{x^2}{3} - \frac{x^2}{3} + 0$$

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

Since the derivative of our result matches the original integrand exactly, the integration is verified as correct.

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 when differentiated)
   Then du = (1/x) dx.

2. Choose dv = x² dx (this is straightforward to integrate)
   Then v = ∫ x² dx = (x³/3).

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

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

5. Combine:
   ∫ x² ln(x) dx = \frac{x^3 \ln x}{3} - \frac{x^3}{9} + C,
   where C is the constant of integration.

   (This can also be written as x³ \left( \frac{\ln x}{3} - \frac{1}{9} \right) + C.)

Verification by Differentiation

Differentiate y = \frac{x^3 \ln x}{3} - \frac{x^3}{9} + C to confirm it equals x² ln(x).

• Compute y':
  y' = \frac{d}{dx} \left[ \frac{x^3 \ln x}{3} \right] - \frac{d}{dx} \left[ \frac{x^3}{9} \right] + 0.

• First term (product rule):
  \frac{d}{dx} [x^3 \ln x] = (x^3)' \ln x + x^3 (\ln x)' = 3x^2 \ln x + x^3 \cdot (1/x) = 3x^2 \ln x + x^2.
  Divide by 3: x^2 \ln x + \frac{x^2}{3}.

• Second term:
  \frac{d}{dx} [x^3/9] = (1/9) \cdot 3x^2 = \frac{x^2}{3}.

• Combine:
  y' = \left( x^2 \ln x + \frac{x^2}{3} \right) - \frac{x^2}{3} = x^2 \ln x.

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