# How do you integrate by substitution int 1/sqrt(2x)dx?

Put $y = \sqrt{2 x}$
$x = {y}^{2} / 2$
$\mathrm{dx} = y \mathrm{dy}$
$\int \frac{1}{\sqrt{2 x}} \mathrm{dx} = \int \frac{y}{y} \mathrm{dy} = y + c = \sqrt{2 x} + c$
You can do it without substitution remembering intx^ndx=x^(n+1)/(n+1)+c;\ \ n=-1/2