Hi,
I have to find the integral of 1/square root(x-3)dx
this is what i did
let u=sqr root(x-3)
du=1/2*square root(x-3)dx
dx=2*square root(x-3)du
the integral of 1/square root(x-3)dx= 2*square root(x-3)(1/u)du
..ok now what i did is plug back u= square root of(x-3) in the denom...