Answer
$$2e$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} \cr
& {\text{Using properties of limits}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} \int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{\mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right)}} \cr
& {\text{Evaluate the limits}} \cr
& \frac{{\mathop {\lim }\limits_{x \to 1} \int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{\mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right)}} = \frac{{\int_1^1 {{e^{{t^3}}}dt} }}{{1 - 1}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's Rule}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} \frac{d}{{dx}}\left[ {\int_1^{{x^2}} {{e^{{t^3}}}dt} } \right]}}{{\mathop {\lim }\limits_{x \to 1} \frac{d}{{dx}}\left[ {x - 1} \right]}} \cr
& {\text{Differentiate, recall that }}\frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( {g\left( x \right)} \right)} dt} \right] = f\left( {g\left( x \right)} \right)g'\left( x \right) \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} {e^{{{\left( {{x^2}} \right)}^3}}}\frac{d}{{dx}}\left[ {{x^2}} \right]}}{{\mathop {\lim }\limits_{x \to 1} \left( 1 \right)}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} 2x{e^{{x^6}}}}}{{\mathop {\lim }\limits_{x \to 1} \left( 1 \right)}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} = \frac{{2\left( 1 \right){e^{{{\left( 1 \right)}^6}}}}}{1} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{\int_1^{{x^2}} {{e^{{t^3}}}dt} }}{{x - 1}} = 2e \cr} $$
