If f(x)=0x(t3+3t2+4)dt\displaystyle {f{{\left({x}\right)}}}={\int_{{0}}^{{x}}}{\left({t}^{{3}}+{3}{t}^{{2}}+{4}\right)}{\left.{d}{t}\right.}
then
f(x)=\displaystyle {f}{''}{\left({x}\right)}=