Chapter 4 - Applications of the Derivative - Review Exercises - Page 332: 79

$${\tan ^{ - 1}}x + C$$

$$\eqalign{ & \int {\frac{{dx}}{{{x^2} + 1}}} \cr & {\text{by the differentiation formula }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x} \right] = \frac{1}{{{x^2} + 1}} \cr & d\left[ {{{\tan }^{ - 1}}x} \right] = \frac{1}{{{x^2} + 1}}dx \cr & {\text{then}}{\text{, we conclude that}} \cr & \int {\frac{{dx}}{{{x^2} + 1}}} = {\tan ^{ - 1}}x + C \cr} $$