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 compute:

$$du=\frac{1}{x}\,dx$$

and

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

Step 3: Apply integration by parts

$$\int x^2\ln(x)\,dx = uv-\int v\,du$$

Substitute:

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

Simplify inside the integral:

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

Now integrate:

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

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 $x^3\ln(x)$:

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

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

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

Differentiate the second term

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

Add the results

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

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

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

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

If you want, I can also show a quick shortcut for choosing $u$ in integration by parts.

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

1. Integration by Parts

The formula for Integration by Parts is: $$\int u \, dv = uv - \int v \, du$$

Step 1: Choose $u$ and $dv$ Following the LIATE rule (Logarithmic, Inverse Trig, Algebraic, Trig, Exponential), we prioritize the logarithmic function for $u$:

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

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

Step 3: Simplify and integrate $$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2 \, dx$$ $$\int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \frac{1}{3} \left( \frac{x^3}{3} \right) + C$$

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

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2. Verification by Differentiation

To verify, we differentiate our result $y = \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$ with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) \right] - \frac{d}{dx} \left[ \frac{x^3}{9} \right] + \frac{d}{dx} [C]$$

Step 1: Apply the Product Rule to the first term Using $(fg)' = f'g + fg'$: $$\frac{d}{dx} \left[ \frac{x^3}{3} \ln(x) \right] = \left( \frac{3x^2}{3} \right) \ln(x) + \frac{x^3}{3} \left( \frac{1}{x} \right) = x^2 \ln(x) + \frac{x^2}{3}$$

Step 2: Differentiate the remaining terms $$\frac{d}{dx} \left[ \frac{x^3}{9} \right] = \frac{3x^2}{9} = \frac{x^2}{3}$$ $$\frac{d}{dx} [C] = 0$$

Step 3: Combine the results $$\frac{dy}{dx} = \left( x^2 \ln(x) + \frac{x^2}{3} \right) - \frac{x^2}{3}$$ $$\frac{dy}{dx} = x^2 \ln(x)$$

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