Question

已知导数为1/√x,则原函数为

Answer 1

y' = 1/√x

dy = dx / √x = x^(-1/2) dx

y = ∫ x^(-1/2) dx = x^(-1/2 + 1)/(1/2) + C

= 2x^(1/2) + C

= 2 √x + C

原函数：y(x) = 2√x + C