How Can You Integrate 1/(sqrt(x)-x)?

[beanryu]

Homework Statement

integrate 1/(sqrt(x)-x)

Homework Equations

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...

 

[rocomath]

is your problem?

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

 

[beanryu]

yeah

 

[morphism]

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...

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.

 

[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.

 

[transgalactic]

is the answer :

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

 

[CompuChip]

transgalactic = beanryu?

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

 

[transgalactic]

i tried the method of
x=v^2

it gives me that answer

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

beanryu is it the right answer??

 

[PowerIso]

i tried the method of
x=v^2

it gives me that answer

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

beanryu is it the right answer??

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

 

[transgalactic]

unfortunately no


but where i did wrong

{ is integral sign

i
made x=v^2 >>> dx=2v*dv



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)

 

[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.

 

[transgalactic]

its -ln(1-v)
right??

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

 

[CompuChip]

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

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

 

[transgalactic]

aaaaaaaahhhhhhhhh
damn little mistakes

 

[beanryu]

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

 

[beanryu]

i got the answer,
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... weird...

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~

 

[transgalactic]

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