# Thread: intergrating of rooted function

1. ## intergrating of rooted function

Need help integrating $\frac{\sqrt(x) - 4}{\sqrt(x) - 2}$

2. Originally Posted by foreg0ne
Need help deriving $\frac{\sqrt(x) - 4}{\sqrt(x) - 2}$
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.

$\frac{\sqrt{x} - 4}{\sqrt{x} -2} = \frac{\sqrt{x} - 4}{\sqrt{x} -2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2}$

$= \frac{x - 2\sqrt{x} - 8}{x - 4}$.

Does this help?

3. oops im sorry, i meant integrate (changed the title but forgot the text)
-
rationalizing makes sense...
can i solve the rest by make .../(x-4) = ...(x-4)^-1 and integrate from there?

4. Hello,

I give you the answer, try to find the method :

$\int \frac{\sqrt{x}-4}{\sqrt{x}-2}=x-4\sqrt{x}-8\ln(\sqrt{x}-2)+C$

5. hmm this is what i can think of:

$\int \frac{\sqrt{x}-4}{\sqrt{x}-2} = u/v
= \int ((v^2) ?? -4*2(v')) dx$

i dont know how the root(x) comes in

6. Hi

$\int \frac{\sqrt{x}-4}{\sqrt{x}-2}\:dx=\int \left(1-2\frac{1}{\sqrt{x}-2}\right)\:dx = x-2\int \frac{dx}{\sqrt{x}-2}$

Set $u = \sqrt x$ to calculate the right-hand integral

7. I would be inclined to make the substitution $u= \sqrt{x}= x^{1/2}$ immediately so that $du= (1/2)x^{-1/2}dx= (1/2)(1/u)dx$ and 2u du= dx. The integral becomes $\int\frac{\sqrt{x}-4}{\sqrt{x}- 2}dx= 2\int u\frac{u-4}{u-2}du$

8. are u guys leaving out steps or anything, because i dont follow the progression of ur working out

9. Originally Posted by HallsofIvy
I would be inclined to make the substitution $u= \sqrt{x}= x^{1/2}$ so that $du= (1/2)x^{-1/2}dx= (1/2)(1/u)dx$ and 2u du= dx. The integral becomes $\int\frac{\sqrt{x}-4}{\sqrt{x}- 2}dx= 2\int u\frac{u-4}{u-2}du$
$2\int \frac{u^2- 4u}{u-2}du= 2\int\left(u- 2- \frac{4}{u-2}\right)du$