Integral of 1/(sqrt(x)-x)

1. Oct 7, 2007

beanryu

1. The problem statement, all variables and given/known data
integrate 1/(sqrt(x)-x)

2. Relevant equations

3. The attempt at a solution

1st I tried to do it the "regular" way, I know that it must be the derivative of

ln(sqrt(x)-x), but the derivative of sqrt(x)-x is not 1. So I added stuff to the equation to make it like 1/(sqrt(x)-x) = ((1/2)(x)^(-1/2)-(1/2)(x)^(-1/2)-1+2)/(sqrt(x)-x)... and more things goes on and I got something like (4*sqrt(x)-1)/(2*(x-x*sqrt(x)))... which is obviously not the answer.

2nd I tried to use t = sqrt(x) and 1/(sqrt(x)-x) = dt^2/(t-t^2), but I have no idea how to solve this....

2. Oct 8, 2007

rocomath

$$\int\frac{dx}{\sqrt{x}-x}$$

3. Oct 8, 2007

yeah

4. Oct 8, 2007

morphism

If t=sqrt(x), then dt/dx=1/(2sqrt(x))=1/(2t). So your integrand should become 2t/(t-t^2). It should be pretty easy to take it from here.

5. Oct 8, 2007

Avodyne

When you have the square root of something in an integral, it's offen helpful to make the substitution u2 = thing that's under the square root sign.

6. Oct 8, 2007

transgalactic

2* ln(1-x^2) +c

7. Oct 8, 2007

CompuChip

transgalactic = beanryu?

And did you try differentiating it and check what comes out?

8. Oct 8, 2007

transgalactic

i tried the method of
x=v^2

2* ln(1-x^2) +c

beanryu is it the right answer??

Last edited: Oct 8, 2007
9. Oct 8, 2007

PowerIso

Take the derivative and if you get the same thing that is in your integral, you are right.

10. Oct 8, 2007

transgalactic

unfortunatly no

but where i did wrong

{ is integral sign

i

then it gives us

{ (2v*dv)/(v-v^2) that gives us

{ 2*dv/(1-v) = 2*ln(1-v) = 2*ln(1-x^2)

Last edited: Oct 8, 2007
11. Oct 8, 2007

Avodyne

1) The integral of dv/(1-v) is NOT ln(1-v). (Close, but not quite. Hint: if you define u=v-1, what do you get?)

2) v is NOT equal to x^2.

12. Oct 8, 2007

transgalactic

its -ln(1-v)
right??

what do meen v is not equal to x^2
i defined that

13. Oct 8, 2007

CompuChip

No, you defined v^2 = x.
What does that make v?

(By the way, you're correct on the -ln(1-v) now)

14. Oct 8, 2007

transgalactic

aaaaaaaahhhhhhhhh
damn little mistakes

15. Oct 8, 2007

beanryu

if x=v^2 >>> shouldn't dx = d(v^2)?

16. Oct 8, 2007

beanryu

my answer is (-2)ln(sqrt(x)-1) using u^2 = x method,

than the answer in the textbook is (-2)ln(1-sqrt(x)),

I took both derivativea they are the same.... wierd....

if dx/(sqrt(x)-x) = dy/(tan(y)), should I use answer (-2)*ln(1-sqrt(x)), because this is the function of x when x<1.

thank you guys~

Last edited: Oct 8, 2007
17. Oct 9, 2007

transgalactic

dx = d(v^2)=2v*dv