Chapter 4 - Integration - 4.6 The Fundamental Theorem Of Calculus - Exercises Set 4.6 - Page 321: 50

See explanation.

$(a)$ $\frac{d}{d x} \int_{0}^{x} \frac{d t}{\sqrt{t}+1}=\frac{1}{\sqrt{x}+1}$ $(\mathrm{b})$ \[ \frac{d}{d x} \int_{1}^{x} \ln t d t=\ln x \] Remember : \[ \frac{d}{d x} \int_{a}^{x} f(t) d t=f(x) \] So: $$=\frac{1}{x^2-3x-4}$$