How do I integrate 1/(1 + x²) and why does it give arctan?

This is a standard result that comes from reversing a known derivative. If y = arctan x, that is tan inverse of x, then by differentiating you get dy/dx = 1/(1 + x²). Since integration is the reverse of differentiation, the integral of 1/(1 + x²) dx must be arctan x plus a constant C. You can see why by substitution: let x = tan θ, so dx = sec²θ dθ and 1 + x² = sec²θ. The integral becomes the integral of sec²θ/sec²θ dθ, which is just the integral of dθ = θ, and θ = arctan x. A useful extension is that the integral of 1/(a² + x²) dx equals (1/a) arctan(x/a) + C, which appears often in board exams.