To solve
$$\int x^2 \ln(x)\,dx,$$ we use integration by parts.

Step 1: Recall the formula

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

Step 2: Choose $u$ and $dv$

Let $$u=\ln(x), \qquad dv=x^2\,dx$$

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

Step 3: Apply integration by parts

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

Simplify inside the integral: $$= \frac{x^3\ln(x)}{3} - \frac{1}{3}\int x^2\,dx$$

Step 4: Integrate the remaining term

$$\int x^2\,dx=\frac{x^3}{3}$$

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

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

Final Answer

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

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Verification by differentiation

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

Differentiate the first term

Use the product rule on $\frac{1}{3}x^3\ln(x)$: $$\frac{d}{dx}\left(\frac{x^3\ln(x)}{3}\right) = \frac{1}{3}\left(3x^2\ln(x)+x^3\cdot \frac{1}{x}\right)$$ $$= x^2\ln(x)+\frac{x^2}{3}$$

Differentiate the second term

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

Add them

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

This matches the original integrand, so the answer is correct.

$$\boxed{\frac{d}{dx}\left(\frac{x^3\ln(x)}{3}-\frac{x^3}{9}\right)=x^2\ln(x)}$$

Step 1: Identify the integration technique

This integral requires integration by parts: $$\int u \, dv = uv - \int v \, du$$

Step 2: Choose $u$ and $dv$

For $\int x^2 \ln(x) , dx$, we choose:

• $u = \ln(x)$  (because it simplifies when differentiated)
• $dv = x^2 , dx$  (because it's easy to integrate)

Step 3: Calculate $du$ and $v$

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

Step 4: Apply the integration by parts formula

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

Step 5: Simplify the remaining integral

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

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

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

Step 6: Final answer

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

Or factored: $\displaystyle \frac{x^3}{9}(3\ln(x) - 1) + C$

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Verification by Differentiation

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

Differentiate term by term:

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

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

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

Combine: $$F'(x) = x^2\ln(x) + \frac{x^2}{3} - \frac{x^2}{3} + 0 = \boxed{x^2\ln(x)} \quad \checkmark$$

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