f(x) = ln x / sqrtx

==> u= ln x ==> du = 1/x dx

==> dv = (1/sqrtx) ==> v = 2sqrtx

==> Int u dv = u*v - INt v du

= 2sqrtx*ln x - Int (2sqrtx/x) dx

= 2sqrtx*lnx - 2*Int 1/sqrtx dx

= 2sqrtx*lnx - 2*2sqrtx

= 2sqrtx*lnx - 4 sqrtx

= 2sqrtx(lnx -2)

(((To prove tthat Int 1/sqrtx dx = 2sqrtx :

Let u= sqrtx ==> du = (1/2sqrtx) dx

==> du = 1/2u dx

==> 2u du = dx

==> Int 1/sqrtx dx = Int 1/u * 2u du = Int 2 du = 2u = 2sqrtx ))))

**==> Int lnx/sqrtx = 2sqrtx*(lnx -2).**

To determine the antiderivative of function, we'll calculate the indefinite integral of function. We'll use integration by parts.

The formula of integration by parts is:

Int udv = u*v - Int vdu

We'll put u = ln x => du = dx/x

We'll put v' = 1/sqrt x => v = 2sqrtx

We'll substitute u,v,u',v' into the formula above:

Int udv = 2sqrtx*ln x - Int 2sqrtx dx/x

Int udv = 2sqrtx*ln x - Int 2dx/sqrtx

But Int dx/sqrtx = Int (2sqrt x)' = 2*2sqrt x

Int udv = 2sqrtx*ln x - 4sqrt x + C

Int udv = 2sqrtx*(ln x - 2) + C

**The antiderivative of the given function f(x) = ln x/sqrtx is F(x) = 2sqrtx*(ln x - 2) + C.**