If ∫ _ 0 ^ 1 ( x ^ 21 + x ^ 14 + x ^ 7 ) (2 x ^ 14 + 3 x ^ 7 + 6) ^...

Let

$I = ∫ \left({x}^{20} + {x}^{13} + {x}^{6}\right)\left(2{x}^{21} + 3{x}^{14} + 6{x}^{7}\right)\dfrac{1}{7}dx$

Put

$2{x}^{21} + 3{x}^{14} + 6{x}^{7} = t$

$⇒ 42\left({x}^{20} + {x}^{13} + {x}^{6}\right)dx = dt$

$∴ I = \dfrac{1}{42}{∫}_{0}^{11}t\dfrac{1}{7}dt$

$⇒ I = \dfrac{1}{42}{\left[\dfrac{7t\dfrac{8}{7}}{8}\right]}_{0}^{11}$

$⇒ I = \dfrac{1}{48}\left(11\right)\dfrac{8}{7}$

So,

$l = 48, m = 8, n = 7$

$∴ l + m + n = 63$