∫dxa2−b2x23/2 equals
xa2−b2x2+C
xa2a2−b2x2+C
axa2−b2x2+C
1a2a2−b2x2+C
Put x=absinθ⇒dx=abcosθdθ∫dxa2−b2x23/2=ab∫cosθdθa3cos3θ=1a2btanθ+C =1a2bsinθ1−sin2θ+C =1a2bbxa1−b2a2x2+C=1a2xa2−b2x2+C