Using Part 1 of the Fundamental Theorem of Calculus

(a) \(\dfrac{d}{dx}\int_{0}^{x}\sqrt{t+1}\,dt=\sqrt{ x+1}\)

(b) \(\dfrac{d}{dx}\int_{2}^{x}\dfrac{s^{3}-1}{2s^{2}+s+1}ds=\dfrac{x^{3}-1}{2x^{2}+x+1}\)