_ x → 0 48 x ^ 4 ∫ _ 0 ^ x t ^ 3 t ^ 6 + 1 dt is equal to

Let

$L = \lim_{x \to 0}\left\{\dfrac{48{∫}_{0}^{x}\left(\dfrac{{t}^{3}}{1 + {t}^{6}}\right)dt}{{x}^{4}}\right\} \to \dfrac{0}{0}$

Applying L'hospital rule, we get

$L = \lim_{x \to 0}\left\{\dfrac{\left(\dfrac{48{x}^{3}}{1 + {x}^{6}}\right)}{4{x}^{3}}\right\}$

$⇒ L = \lim_{x \to 0}\left(\dfrac{12}{1 + {x}^{6}}\right)$

$⇒ L = 12$