Answer: [tex]\large \boldsymbol{\dfrac{1}{8 } (x^2-3)^4+c}[/tex]
Step-by-step explanation:
[tex]\displaystyle \large \boldsymbol{} \int\limits {x(x^2-3)^2 } \, dx =\int\limits \frac{1}{2} {} \, (x^2-3)' (x^2-3)^3 dx =\frac{1}{8} (x^2-3)^4+c[/tex]
