Question

不定积分证明∫√x∧2+a∧2

Answer 1

令x = a * tanθ
∫ √(x² + a²) dx
= ∫ √(a²tan²θ + a²) d(a * tanθ)
= a²∫ secθ dtanθ
= a²secθtanθ - a²∫ tanθ dsecθ
= a²secθtanθ - a²∫ tan²θsecθ dθ
= a²secθtanθ - a²∫ (sec²θ - 1)secθ dθ
= a²secθtanθ - a²∫ sec³θ dθ + a²∫ secθ dθ
2a²∫ sec³θ dθ = a²secθtanθ + a²ln|secθ + tanθ|
a²∫ sec³θ dθ = (a²/2)[x/a * √(x² + a²)/a] + (a²/2)ln|x/a + √(x² + a²)/a| + C
==> ∫ √(x² + a²) dx = (x/2)√(x² + a²) + (a²/2)ln|x + √(x² + a²)| + C