#### Answer

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

#### Work Step by Step

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