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

beanryu

Homework Statement

integrate 1/(sqrt(x)-x)

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

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

beanryu
yeah

Homework Helper
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.

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

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

Homework Helper
transgalactic = beanryu?

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

transgalactic
i tried the method of
x=v^2

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

beanryu is it the right answer??

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PowerIso
i tried the method of
x=v^2

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
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)

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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

Homework Helper
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