Question

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

Answer 1

Put `y=sqrt(2x)`
`x=y^2/2`
`dx=ydy`
`int1/sqrt(2x)dx=inty/ydy=y+c=sqrt(2x)+c`

Explanation:

You can do it without substitution remembering `intx^ndx=x^(n+1)/(n+1)+c;\ \ n=-1/2`