We need to compute

$$\int x^2 \ln(x)\,dx$$

using integration by parts.

Step 1: 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 2: Apply integration by parts

Recall:

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

So,

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

Simplify inside the remaining integral:

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

Step 3: Integrate the remaining term

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

Therefore,

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

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

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

$$\frac{d}{dx}\left(\frac{x^3\ln(x)}{3}\right) = \frac{1}{3}\frac{d}{dx}(x^3\ln(x))$$

Use the product rule:

$$\frac{d}{dx}(x^3\ln(x)) = 3x^2\ln(x)+x^3\cdot\frac{1}{x}$$

$$=3x^2\ln(x)+x^2$$

So,

$$\frac{d}{dx}\left(\frac{x^3\ln(x)}{3}\right) = \frac{1}{3}\left(3x^2\ln(x)+x^2\right) = x^2\ln(x)+\frac{x^2}{3}$$

Differentiate the second term

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

Add the derivatives

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

$$F'(x)=x^2\ln(x)$$

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

$$\boxed{F'(x)=x^2\ln(x)}$$

Solving ∫x²·ln(x) dx

Setting Up Integration by Parts

Formula: ∫u·dv = u·v - ∫v·du

Choosing u and dv:

Choice	Expression	Reason
u = ln(x)	→ du = (1/x)dx	Differentiates to simpler form
dv = x²dx	→ v = x³/3	Easy to integrate

Why this assignment? ln(x) has no clean antiderivative, so it must be u, not dv.

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Applying the Formula

$$\int x^2 \ln(x)\, dx = \underbrace{\ln(x)}_{u} \cdot \underbrace{\frac{x^3}{3}}_{v} - \int \underbrace{\frac{x^3}{3}}_{v} \cdot \underbrace{\frac{1}{x}\,dx}_{du}$$

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

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

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

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

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Verification by Differentiation

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

Term 1 — using the Product Rule on $\frac{x^3 \ln(x)}{3}$:

$$\frac{d}{dx}\left[\frac{x^3 \ln(x)}{3}\right] = \frac{1}{3}\left[3x^2 \cdot \ln(x) + x^3 \cdot \frac{1}{x}\right] = x^2\ln(x) + \frac{x^2}{3}$$

Term 2:

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

Combining:

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

The derivative exactly recovers the original integrand, confirming the answer is correct.