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# Evaluate the integral of (2 – x – x2)1/2 / x2

Compute the following integral. Following Apostol’s hint we multiply the numerator and denominator by , Now, we’ll evaluate these three integrals separately. For the first integral, we make the subsitution This gives us Therefore, we have For the second integral… Incomplete.

For the third and final integral in our expression we have Now, we make the substitution , . This gives us, This is incomplete. Otherwise, I’ll try to get back to it as soon as I figure out how to evaluate the second integral above…

1. tom says:

I got closer using a simpler method, but not 100% sure.Do what Apostol suggested, then, bring a x^2 into the denominator of the square root, making the integral (2-x-x^2)/(x^3*sqrt(2(x^2)-1/x-1)). The derivative under the sqrt is -4/x^3+1/x^2. This we have but two more integrals are created- one resolves to the arcsin portion and the other ends up being integral of 1/(x*sqrt(ax^2+bx+c)). It seems this one is solvable by parts but this is where I’m stuck. It is of the correct form in the integral tables.

2. tom says:

I got very close. Multiply the first integral by 1/2, the second one then adds 1/2x instead of 2. The second integral now has an x instead of x^2 in the denominator. Complete the square, substitute with u, then substitute sin(v) for u. Integrand becomes 1/(3cosv-1). Then use tan half angle substitution which results in a messy (for me) log function. My result was quite close but ends up with the 1/2 factor on the first integral and the log was different but I probably just didn’t reduce right.

• RoRi says:

That’s an interesting idea. I worked on a bit, but can’t get it to reduce correctly either. I’ll work on it some more later… It seems like it should work, but I don’t know how we’ll ever get rid of that extra factor of 1/2 in the first integral.

• tom says:

Yeah, using the 3 integrals but with the 1/2 factor works seems to logical until the end: the middle log term is different then Apostol’s and the arcsin factor gets the 1/2 factor. I did notice someone on math.stackexchange worked this one using an easier route but believe he also had trouble matching the answer.

• RoRi says:

Here’s the link to the question on math.SE for anyone reading this later:
Apostol Exercise 6.25.40 on Math.SE

The comments there seem to suggest the answer in Apostol is wrong. I’ll see if I can confirm that and then update the post with the correct solution and derivation.

• A lie says:

Your substitution at the very beginning is missing a factor of 2 ;du=(x-4)/(2[√2-x-x^2]x^2)