evaluate ((lnx)^2)^ 1/2 dx

I suppose that you need to find the integral of the
function `sqrt((lnx)^2) `  such that:

`int sqrt((lnx)^2)
dx`

You need to remember that `sqrt(x^2) = |x|, `  hence
`sqrt((lnx)^2) = |ln x|`

`` `int sqrt((lnx)^2) dx = int |ln
x| dx`

You need to use integration by parts, hence you
should write the formula such that:

`int udv = uv - int
vdu`

You should consider `u = ln x =gt du = (dx)/x` ,
hence`dv = dx =gt v = x` Substituting `ln x`  for u, `(dx)/x`  for du, x for v and dx
for dv yields:

int |ln x| dx = x*|ln x| - int
x*(dx)/x

`int |ln x| dx = x*|ln x| - int
dx`

`int |ln x| dx = x*|ln x| - x + c =gt int |ln x| dx =
x*(|ln x| - 1) + ` c

`int |ln x| dx = x*(|ln x| - ln e) +
c`

`int |ln x| dx = x*|ln(x/e)| +
c`

Hence, evaluating the integral of function
yields `int sqrt((lnx)^2) dx = x*|ln(x/e)| + c` .