Answer: \(\frac{\mathrm dy}{\mathrm dx}=x^2+3x^2\ln x\); \(\int_1^2 3x^2\ln x\,\mathrm dx=\ln256-\frac73\).

Set up the integral using the correct limits or antiderivative first. When definite integration is used, evaluate the upper limit and lower limit separately before subtracting.

Differentiate \(y=x^3\ln x\) using the product rule:

\(\frac{\mathrm dy}{\mathrm dx}=3x^2\ln x+x^3\cdot\frac1x=x^2+3x^2\ln x.\)

Rearrange this result:

\(3x^2\ln x=\frac{\mathrm d}{\mathrm dx}(x^3\ln x)-x^2.\)

Therefore

\(\int 3x^2\ln x\,\mathrm dx=x^3\ln x-\frac{x^3}{3}+c.\)

Now apply the limits:

\(\int_1^2 3x^2\ln x\,\mathrm dx=\left[x^3\ln x-\frac{x^3}{3}\right]_1^2.\)

This is

\(\left(8\ln2-\frac83\right)-\left(0-\frac13\right)=8\ln2-\frac73.\)

Since \(8\ln2=\ln(2^8)=\ln256\), the answer is \(\ln256-\frac73\).

This gives the required answer: \(\frac{\mathrm dy}{\mathrm dx}=x^2+3x^2\ln x\); \(\int_1^2 3x^2\ln x\,\mathrm dx=\ln256-\frac73\).