In the realm of calculus, certain integrals stand out for their frequent application and the elegant techniques required for their resolution. One such integral is \( \int \dfrac{1}{x^2+1} dx \), notable for its direct connection to the arctangent function, a fundamental element in trigonometric calculations. The process of integrating \( \dfrac{1}{x^2+1} \) showcases the synergy between algebraic expressions and trigonometric identities, simplifying complex calculations elegantly.
\( \int \dfrac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C \)
Solution:
Let \( x = \tan(\theta) \) ⇒ \( dx = \sec^2(\theta) d\theta \) & \( \theta = \tan^{-1}(x) \)
\( \therefore \int \frac{1}{x^2+1} \, dx = \int \frac{1}{\tan^2(\theta) + 1} \sec^2(\theta) \, d\theta \)
= \( \int \sec^2(\theta) \, d\theta \)
= \( \int d\theta \)
= \( \theta + C \)
= \( \tan^{-1}(x) + C \)
