What Is the Correct Substitution for This Integral?

[ascky]

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

[tex]\int \sqrt{\frac{x-1}{x(x+1)}} dx [/tex]

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

 

[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.

 

[ascky]

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

[tex]\int \frac{\sqrt{\frac{x-1}{x+1}}}{x^2} dx [/tex]

Wolfram thinks it's arctan-ny.

 

[morson]

Make the substitution [tex]x = sin^2h(t)[/tex], and the integral reduces to:

[tex]2\int \sqrt{sin^2h(t) - 1}\ dt[/tex]

Edit: just saw your post. Oops.

 

[D H]

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
[tex]\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)[/tex]

 

[morson]

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

So the integral becomes:

[tex]\int \frac{4u^2 du}{(1+u^2)^2}\[/tex]

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: [tex]\frac{4}{1+u^2}\ - \frac{4}{(1+u^2)^2}\[/tex]

 

[ascky]

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

[tex]\int \frac{4u^2}{(1+u^2)^2} du[/tex]

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

 

[morson]

[tex]u^2 = \frac{x-1}{x+1}\ [/tex]
[tex] = 1 - \frac{2}{x+1}\ [/tex]

Solving for (x + 1) gives:

[tex] x + 1 = \frac{2}{1 - u^2}\ [/tex] --- (1)

So, [tex](x + 1)^2 = \frac{4}{(1 - u^2)^2}\ [/tex] --- (2)

Also, [tex]2u du = \frac{2 dx}{(x+1)^2}\ [/tex]

^ You got that part right? So:

[tex]dx = u(x + 1)^2 du[/tex]

From (2),

[tex]dx = \frac{4u du}{(1 - u^2)^2}\ [/tex]

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

[tex] x = \frac{2 - (1 - u^2)}{1 - u^2}\ [/tex]

[tex] = \frac{1 + u^2}{1 - u^2}\ [/tex]

Therefore, [tex]x^2 = \frac{(1 + u^2)^2}{(1 - u^2)^2}\ [/tex]

So, the whole integral becomes:

[tex] \int \frac{ \frac{4u^2 du}{(1 - u^2)^2} }{ \frac{(1 + u^2)^2}{(1 - u^2)^2}\ } \ [/tex]

= [tex] \int \frac{4u^2 du}{(1 + u^2)^2}\ [/tex]

 

[ascky]

Ahhh I get it now. That was very clear.

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

 

[i.mehrzad]

well i tried it by substituting the following:

[tex]x=\tan^2\t[/tex] and then continued. It simplifies very well.

 

[i.mehrzad]

I never get my latex coding right.
[tex]x/=/tan^2t[/tex]

 

[i.mehrzad]

well isn't the backslash to be for space,

[tex]x\=\tan^2\t[/tex]

 

[i.mehrzad]

now i got it all wrong
[tex]x=\tan^2t[/tex]

 

[i.mehrzad]

Now it is fine, sorry for my disobedience with so many posts.