Answer
$$\frac{\partial}{\partial \alpha}\int_\alpha^\beta\sqrt{t^3+1}dt=-\sqrt{\alpha^3+1};$$
$$\frac{\partial}{\partial \beta}\int_\alpha^\beta\sqrt{t^3+1}dt=\sqrt{\beta^3+1}.$$
Work Step by Step
We will use the fact that
$$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$
if $a$ is a constant.
Regard $\beta$ as constant and differentiate with respect to $\alpha$:
$$\frac{\partial}{\partial \alpha}\int_\alpha^\beta\sqrt{t^3+1}dt=\frac{\partial}{\partial \alpha}\left(-\int_\beta^\alpha\sqrt{t^3+1}dt\right)=-\frac{\partial}{\partial \alpha}\left(\int_\beta^\alpha\sqrt{t^3+1}dt\right)=-\sqrt{\alpha^3+1}.$$
Regard $\alpha$ as constant and differentiate with respect to $\beta$:
$$\frac{\partial}{\partial \beta}\int_\alpha^\beta\sqrt{t^3+1}dt=\sqrt{\beta^3+1}.$$
