Question

If $\int\limits_0^1\left(x^{21}+x^{14}+x^7\right)\left(2 x^{14}+3 x^7+6\right)^{1 / 7} d x=\frac{1}{l}(11)^{m / n}$ where $l, m, n \in N , m$ and $n$ are coprime then $l+m+n$ is equal to ______

Answer 1

$ \int\left(x^{20}+x^{13}+x^6\right)\left(2 x^{21}+3 x^{14}+6 x^7\right)^{1 / 7} d x$
$ 2 x ^{21}+3 x ^{14}+6 x ^7= t $
$ 42\left( x ^{20}+ x ^{13}+ x ^6\right) dx = dt$
$\frac{1}{42} \int\limits_0^{11} t^{\frac{1}{7}} d t=\left(\frac{t^{\frac{8}{7}}}{\frac{8}{7}} \times \frac{1}{42}\right)_0^{11}$
$ =\frac{1}{48}\left(t^{\frac{8}{7}}\right)_0^{11}=\frac{1}{48}(11)^{8 / 7} $
$ l=48, m =8, n =7 $
$ l+ m + n =63$