Solve Partial Fractions: 1+u2/(1+u4)

[KLscilevothma]

I have been doing revision on integration lately, and came across a question that need to resolve (1+u²)/(1+u⁴) in partial fraction before I can proceed.

1+u⁴ = (u²-2u+2)(x²+2u+2)

therefore 1+u² = (Au+B)(x²+2u+2) + (Cu+D)(u²-2u+2)
and I need to find out A, B, C and D

But wait, the next step in my textbook contains something like (1+sqrt(2)u+u²) and (1-sqrt(2)+u²) in denominators, so what is the problem and where does sqrt(2) come from ?

PS
original question:
[inte]dx/[(1+x²)(sqrt(1-x²))]
and the substitution used is :
u=sqrt[(1-x)/(1+x)]

 

[Hurkyl]

1+u4 = (u2-2u+2)(x2+2u+2)

This is false. If you expand it, you get 4 for your constant term, not 1. I presume that if you replaced your factors with the two with sqrt(2) in them you'd get the correct result.



P.S. why that substitution? It seems to work, but I've never seen that sort of substitution suggested before.

P.P.S. After you get the solution your way, try substituting x=sin θ and compare the work done.

 

[KLscilevothma]

Thanks for your reply.

1+u4 = (u^2-2u+2)(x^2+2u+2)

This is false

Oh yes, though I've expanded the brackets and checked, thought 4 was the constant instead of 1. I also thought
cos(pi/4)+isin(pi/4)=1+i, that's why I got it wrong. (I did my revision for more than 4 hours till 4am, and it was one of the last questions I did)

why that substitution

According to my book, it's called Euler's transformation. If we want to evaluate integrals of the type [squ](ax²+bx+c) and if ax²+bx+c can be factorized into real linear factors a(x-[alpha])(x-[beta]), use
u=[squ][a(x-[alpha])/(x-[beta])] as substitution.

For example, in this question,
[inte]dx/{[(1+x²)sqrt(1-x²)]}
=[inte]dx/{(1+x²)(1+x)sqrt[(1-x)(1+x)]}

substitution, u=sqrt[(1-x)/(1+x)]

In another textbook, t² = (1-x²)(1+x²) is used as substitution. It seems that the second method is easier and doesn't need to use partial fraction.