Answer: \(\dfrac{dy}{dx}=x+2x\ln x\); \(\displaystyle\int x\ln x\,dx=\dfrac12x^2\ln x-\dfrac{x^2}{4}+C\).

We start with the main method. Integrate term by term and include the constant of integration for an indefinite integral.

Given

\(y=x^2\ln x.\)

Differentiate using the product rule:

\(\frac{dy}{dx}=x^2\cdot\frac1x+\ln x\cdot2x.\)

So

\(\frac{dy}{dx}=x+2x\ln x.\)

For part (b), rearrange this derivative:

\(\frac{d}{dx}(x^2\ln x)=x+2x\ln x.\)

Therefore

\(2x\ln x=\frac{d}{dx}(x^2\ln x)-x.\)

Divide by \(2\):

\(x\ln x=\frac12\frac{d}{dx}(x^2\ln x)-\frac{x}{2}.\)

Now integrate both sides:

\(\int x\ln x\,dx=\frac12x^2\ln x-\int\frac{x}{2}\,dx.\)

Hence

\(\int x\ln x\,dx=\frac12x^2\ln x-\frac{x^2}{4}+C.\)

This completes the solution and gives the required result.