Can't figure out this integral

1. Jun 20, 2007

ascky

I've been trying to figure this out for a while, but I can't find the right substitution:

$$\int \sqrt{\frac{x-1}{x(x+1)}} dx$$

Anyone able to help me out? It's driving me crazy.

2. Jun 20, 2007

Pseudo Statistic

Well, Wolfram seems to think that it reduces to the sum of an elliptic integral of the first kind and second kind, furthermore involving complex numbers.

3. Jun 20, 2007

ascky

Oh, gee, I copied it wrong. How embarrassing. My apologies. How about

$$\int \frac{\sqrt{\frac{x-1}{x+1}}}{x^2} dx$$

Wolfram thinks it's arctan-ny.

4. Jun 20, 2007

morson

Make the substitution $$x = sin^2h(t)$$, and the integral reduces to:

$$2\int \sqrt{sin^2h(t) - 1}\ dt$$

Edit: just saw your post. Oops.

5. Jun 20, 2007

Staff: Mentor

You get an ugly mess if you type that into Wolfram's Integrator directly. On the other hand, simplifying the radical leads to the much simpler form
$$\int\frac{\sqrt{x^2-1}}{x^2(x+1)}dx = -\left(\tan^{-1}\left(\frac1{\sqrt{x^2-1}}\right)+\frac{\sqrt{x^2-1}}x\right)$$

6. Jun 20, 2007

morson

In that case, make the substitution $$u^2 =\frac{x-1}{x+1}\$$.

So the integral becomes:

$$\int \frac{4u^2 du}{(1+u^2)^2}\$$

Looks like a further substitution of u = tan(t) or u = sinh(a) will do the trick. Or partial fractions.

By partial fractions, the integrand is: $$\frac{4}{1+u^2}\ - \frac{4}{(1+u^2)^2}\$$

Last edited: Jun 20, 2007
7. Jun 20, 2007

ascky

I'm having trouble with the algebra manipulating that substitution to get it nicely as

$$\int \frac{4u^2}{(1+u^2)^2} du$$

Would it be too much trouble to explain how you did that? Of course from there, it's nice.

8. Jun 21, 2007

morson

$$u^2 = \frac{x-1}{x+1}\$$
$$= 1 - \frac{2}{x+1}\$$

Solving for (x + 1) gives:

$$x + 1 = \frac{2}{1 - u^2}\$$ --- (1)

So, $$(x + 1)^2 = \frac{4}{(1 - u^2)^2}\$$ --- (2)

Also, $$2u du = \frac{2 dx}{(x+1)^2}\$$

^ You got that part right? So:

$$dx = u(x + 1)^2 du$$

From (2),

$$dx = \frac{4u du}{(1 - u^2)^2}\$$

So, the expression inside the square root of your original integral becomes u^2, so the whole square root becomes u. From (1):

$$x = \frac{2 - (1 - u^2)}{1 - u^2}\$$

$$= \frac{1 + u^2}{1 - u^2}\$$

Therefore, $$x^2 = \frac{(1 + u^2)^2}{(1 - u^2)^2}\$$

So, the whole integral becomes:

$$\int \frac{ \frac{4u^2 du}{(1 - u^2)^2} }{ \frac{(1 + u^2)^2}{(1 - u^2)^2}\ } \$$

= $$\int \frac{4u^2 du}{(1 + u^2)^2}\$$

9. Jun 21, 2007

ascky

Ahhh I get it now. That was very clear.

Thanks very much for your help. Now I can sleep contentedly!

10. Jun 21, 2007

well i tried it by substituting the following:

$$x=\tan^2\t$$ and then continued. It simplifies very well.

11. Jun 21, 2007

I never get my latex coding right.
$$x/=/tan^2t$$

12. Jun 21, 2007

well isnt the backslash to be for space,

$$x\=\tan^2\t$$

13. Jun 21, 2007

$$x=\tan^2t$$