The value of ${\int }_0^1(x^{12}+x^8+x^4){(2x^8+3x^4+6)}^{\frac{1}{4}}dx=\frac{1}{l}{\left(11\right)}^{\frac{m}{n}},$ where $x>0,$ and m, n are coprime then $l+m+n=$

${\int }_0^1(x^{12}+x^8+x^4){(2x^8+3x^4+6)}^{\frac{1}{4}}dx$

$\underset{0}{\overset{1}{\int }}(x^{11}+x^7+x^3).{(2x^{12}+3x^8+6x^4)}^{\frac{1}{4}}dx$

$Let 2x^{12}+3x^8+6x^4=t\Rightarrow 24\left(x^{11}+x^7+x^3\right)dx=dt$

$=\frac{1}{24}\underset{0}{\overset{1}{\int }}{\left(t\right)}^{\frac{1}{4}}dt=\frac{1}{24}\times \frac{4}{5}{\left\{t^{\frac{5}{4}}\right\}}_0^{11}=\frac{1}{30}{\left(11\right)}^{\frac{5}{4}}$