Đáp án:

• Đặt \( x = \tan t \implies dx = (1 + \tan^2 t) \, dt \)  

• \( \int \frac{1}{1 + x^2} \, dx = \int \frac{1}{1 + \tan^2 t} (1 + \tan^2 t) \, dt = \int dt \)  

                   \( = t + C = \arctan x + C \)