Need help integrating $\displaystyle \frac{\sqrt(x) - 4}{\sqrt(x) - 2}$
Hang on.
Are you finding an integral? A derivative? Or something else?
Integrating is not the same as deriving and neither are the same as simplifying.
Whatever the case, most likely the easiest way to solve your problem will be to simplify it first. To do this, multiply top and bottom by the conjugate of the bottom.
$\displaystyle \frac{\sqrt{x} - 4}{\sqrt{x} -2} = \frac{\sqrt{x} - 4}{\sqrt{x} -2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2}$
$\displaystyle = \frac{x - 2\sqrt{x} - 8}{x - 4}$.
Does this help?
I would be inclined to make the substitution $\displaystyle u= \sqrt{x}= x^{1/2}$ immediately so that $\displaystyle du= (1/2)x^{-1/2}dx= (1/2)(1/u)dx$ and 2u du= dx. The integral becomes $\displaystyle \int\frac{\sqrt{x}-4}{\sqrt{x}- 2}dx= 2\int u\frac{u-4}{u-2}du$