Respuesta :
if y = xⁿ
∫y dx = xⁿ⁺¹ / (n + 1) + C Provided n ≠ -1.
y = √x
y = x^(0.5)
∫y dx = x^(0.5+1) / (0.5 + 1) = x^(1.5) / 1.5 = x^(1.5) / (3/2)
∫y dx = (2/3) x^(1.5) + C.
∫y dx = (2/3) x^(3/2) + C.
∫y dx = (2/3)√x³ + C
∫y dx = xⁿ⁺¹ / (n + 1) + C Provided n ≠ -1.
y = √x
y = x^(0.5)
∫y dx = x^(0.5+1) / (0.5 + 1) = x^(1.5) / 1.5 = x^(1.5) / (3/2)
∫y dx = (2/3) x^(1.5) + C.
∫y dx = (2/3) x^(3/2) + C.
∫y dx = (2/3)√x³ + C
Its not guaranteed to be same form.it could be any polynomial that can not be easily formed into squares .try one out!