Zu berechnen ist:
⟪int_x dx/sqrt(a²+x²)⟫
Wir bereiten eine Substituion vor:
(1)
⟪u = x + sqrt(a²+x²)⟫
(2)
⟪sqrt(a²+x²) = u - x⟫
⟪a² + x² = u² -2ux + x²⟫
⟪2ux = u² - a²⟫
⟪x = u/2 - (a²)/(2u)⟫
(3)
⟪dx = (1/2 + (a²)/(2u²)) du⟫
(4)
⟪sqrt(a²+x²) = u - x = u/2 + (a²)/(2u)⟫
Wir integrieren durch Substitution:
⟪int_x dx/sqrt(a²+x²) = int_u (1/2 + (a²)/(2u²)) / (u/2 + (a²)/(2u)) du = int_u 1/u du = ln u = ln (x + sqrt(a²+x²)) ∎⟫