Expert: Steve Holleran - 9/27/2007

Question
show that the following integral [dx/(x+1)^3/2 + (x-1)^3/2] may be transformed in to the integral of a rational function by the means of the substitution: x= 1/2(t^2 + t^(-2))

Answer
Hi Amery,

Sorry this took so long; I just started a new teaching job, and the startup was crazy.

Anyway, I think I might have come up with something that might be what you need.

If you make the substitutions, in the denominator you get:

[1/2 (t^2 + 1/t^2) +1]^3/2  + [1/2(t^2 + 1/t^2)-1]^3/2

If you get a common denominator in each bracket of 2t^2, you will have :

 [t^4 + 1 + 2t^2 / 2t^2]^3/2 + [t^4 +1 - 2t^2 / 2t^2]^3/2

= [ (t^2+1)^2 / 2t^2]3/2  + [(t^2 -1)^2 / 2t^2]^3/2

=[(t^2+1)^3 / sqrt8 * t^3]  + [ (t^2 - 1)^3 / sqrt8 * t^3]

=  [(t^2 + 1)^3 + (t^2 - 1)^3] / sqrt8 * t^3

Since this is the denominator in the integral, it will invert and the integral will now be :

INT{ (sqrt8 * t^3) dx / [(t^2 + 1)^3 + (t^2 - 1)^3]}

Now, if x = 1/2 *(t^2 + t^-2) then

dx = 1/2 * (2t - 2t^-3) * dt = (t - 1/t^3) dt  

Now in the numerator of the INT, you can pull the sqrt8 out to the front, and the numerator will be :

       t^3 * (t - 1/t^3) * dt = (t^4 - 1) * dt

and now the integral will be :

sqrt8 * INT [ (t^4-1) dt / [(t^2+1)^3 + (t^2-1)^3]]

and this is a rational function.  I don't know what to do with it at this point, but I hope this helps somewhat.

Again, sorry I was so slow.

Steve