Perplexity vs Kimi for Math

To solve $\int x^2 \ln x \, dx$, use integration by parts, with the formula $\int u \, dv = uv - \int v \, du$. Set $u = \ln x$ (so $du = \frac{1}{x} dx$) and $dv = x^2 dx$ (so $v = \frac{x^3}{3}$)[1][2][3].

This gives: $$\int x^2 \ln x \, dx = \ln x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3 \ln x}{3} - \frac{1}{3} \int x^2 \, dx.$$

Integrate the remaining term using the power rule: $$\int x^2 \, dx = \frac{x^3}{3},$$ so $$\frac{x^3 \ln x}{3} - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3 \ln x}{3} - \frac{x^3}{9} + C.$$

The final antiderivative is $\frac{x^3}{3} \ln x - \frac{x^3}{9} + C$[1][2][3].

Verification by differentiation: Differentiate $\frac{x^3}{3} \ln x - \frac{x^3}{9}$ using the product rule on the first term and power rule on the second.

• Product rule: $\frac{d}{dx} \left( \frac{x^3}{3} \ln x \right) = \frac{x^3}{3} \cdot \frac{1}{x} + \ln x \cdot \frac{3x^2}{3} = \frac{x^2}{3} + x^2 \ln x$.
• Power rule: $\frac{d}{dx} \left( -\frac{x^3}{9} \right) = -\frac{3x^2}{9} = -\frac{x^2}{3}$.

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

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

Step 1: Choose $u$ and $dv$

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

Step 2: Apply the formula $$\int x^2\ln(x)\,dx = \frac{x^3}{3}\ln(x) - \int \frac{x^3}{3} \cdot \frac{1}{x}\,dx$$

Step 3: Simplify the remaining integral $$\int \frac{x^3}{3} \cdot \frac{1}{x}\,dx = \frac{1}{3}\int x^2\,dx = \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9}$$

Step 4: Combine terms $$\int x^2\ln(x)\,dx = \frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C$$

This can also be written as: $$\boxed{\int x^2\ln(x)\,dx = \frac{x^3}{9}\left(3\ln(x) - 1\right) + C}$$

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Verification by Differentiation

Let $F(x) = \frac{x^3}{3}\ln(x) - \frac{x^3}{9}$

Differentiate the first term (using the product rule): $$\frac{d}{dx}\left[\frac{x^3}{3}\ln(x)\right] = \frac{3x^2}{3}\cdot\ln(x) + \frac{x^3}{3}\cdot\frac{1}{x} = x^2\ln(x) + \frac{x^2}{3}$$

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

Combine the derivatives: $$F'(x) = x^2\ln(x) + \frac{x^2}{3} - \frac{x^2}{3} = x^2\ln(x)$$

Since $F'(x) = x^2\ln(x)$, which matches the original integrand, the solution is verified.