Perplexity vs Qwen for Math

To solve $\int x^2 \ln x \, dx$, use integration by parts, where $\int u \, dv = uv - \int v \, du$.[1][2][3]

Set $u = \ln x$, so $du = \frac{1}{x} dx$, and $dv = x^2 dx$, so $v = \frac{x^3}{3}$.[1][2][3]

Then, $$\int x^2 \ln x \, dx = \ln x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3 \ln x}{3} - \frac{1}{3} \int x^2 \, dx.$$[1][2][3]

Integrate: $\int x^2 \, dx = \frac{x^3}{3}$, so $$\frac{x^3 \ln x}{3} - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3 \ln x}{3} - \frac{x^3}{9} + C.$$[1][2][3]

Verification by differentiation: Differentiate $\frac{x^3 \ln x}{3} - \frac{x^3}{9}$.

Product rule on first term: $\frac{d}{dx} \left( \frac{x^3 \ln x}{3} \right) = \frac{1}{3} \left( 3x^2 \ln x + x^3 \cdot \frac{1}{x} \right) = x^2 \ln x + \frac{x^2}{3}$.

Second term: $\frac{d}{dx} \left( -\frac{x^3}{9} \right) = -\frac{3x^2}{9} = -\frac{x^2}{3}$.

Total: $x^2 \ln x + \frac{x^2}{3} - \frac{x^2}{3} = x^2 \ln x$, matching the integrand.[1][2][3]

To solve the integral $\int x^2 \ln(x) , dx$, we use the method of Integration by Parts. The formula for Integration by Parts is:

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

Step 1: Assign $u$ and $dv$

Following the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we choose the logarithmic function for $u$ and the algebraic function for $dv$:

• $u = \ln(x) \implies du = \frac{1}{x} , dx$
• $dv = x^2 , dx \implies v = \frac{x^3}{3}$

Step 2: Apply the Formula

Substitute these 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 3: Simplify and Evaluate the Integral

Simplify the expression inside the remaining integral: $$\left( \frac{x^3}{3} \right) \left( \frac{1}{x} \right) = \frac{x^2}{3}$$

Now, integrate the simplified term: $$= \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2 \, dx$$ $$= \frac{x^3}{3} \ln(x) - \frac{1}{3} \left( \frac{x^3}{3} \right) + C$$ $$= \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$$

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

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Verification by Differentiation

To verify the result, we differentiate $F(x) = \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$.

1. Differentiate the first term (Product Rule): $$\frac{d}{dx} \left( \frac{x^3}{3} \ln(x) \right) = \left( \frac{d}{dx} \frac{x^3}{3} \right) \ln(x) + \frac{x^3}{3} \left( \frac{d}{dx} \ln(x) \right)$$ $$= x^2 \ln(x) + \frac{x^3}{3} \left( \frac{1}{x} \right) = x^2 \ln(x) + \frac{x^2}{3}$$

2. Differentiate the second term (Power Rule): $$\frac{d}{dx} \left( -\frac{x^3}{9} \right) = -\frac{3x^2}{9} = -\frac{x^2}{3}$$

3. Combine terms: $$F'(x) = \left( x^2 \ln(x) + \frac{x^2}{3} \right) - \frac{x^2}{3} + 0$$ The terms $+\frac{x^2}{3}$ and $-\frac{x^2}{3}$ cancel out: $$F'(x) = x^2 \ln(x)$$

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