study of a certain integral v2

Study of a certain Integral PhL 11.1.16 Motivation: This appears on page 77 of Goldstein. The integral is the following indefinite integral whose results I quote from GR7 page 94 Here TI is a Russian source I don't have The second DW result seems odd since a does not appear, but you are allowed to drop "constants" in expressing an indefinite integral, so maybe that is where it went! Question: How do you actually manually do such an integral? GR7 gives us a hint as follows, involving Euler substitutions, In my case we have c = -1, and a and b could have either sign, so none of these directly applies, but maybe something similar does. Wiki on these subs refers to a Zwillinger book "The Handbook of Integration" 1992. I have two of Z's books but not this one. I am taking a web look for this book, it does exist and is about integration methods, something I would like to have, http://en.booksee.org/book/1423095 claims to have it, download is slow, we shall see. Yes, I have it! Wiki mentions that above Euler things as well but has different expressions!!! Here they are from wiki 1. = ±x+t 2. = xt ± 3. = (x-α)t where α is a root of R. But wiki have a↔c from GR7, so I change wiki to read 1. = ±x+t 2. = xt ± 3. = (x-α)t where α is a root of R. These are numbered differently from GR7, but they are the same 3 subs! Relation is 1↔2. and they give a good example. Maybe Z will have nothing more than those subs.. // It is a good book, Z knows a lot of stuff. Z born 1957 so is age 43+15 = 58, has some miles to go, I have had some emails with him. The book however does not mention this kind of integral, index does not include Euler subs. Try an Euler substitution Consider: R = (a + bx + cx2) = xt ± (1) t(x) = ( ∓ )/x I want this to have the form x = f(t) and then we can do dx = f'(t) dt So square (1) R = (a + bx + cx2) = ( xt ± )2 = x2t2 ± 2xt + a Notice that the a cancels on both sides, so left with bx + cx2 = x2t2 ± 2xt or b + cx = xt2 ± 2t or x(c-t2) = ± 2t - b x = ( ± 2t - b)(c-t2)-1 Maple confirms where I select the + sign Now want to compute dx : So at this point I have (still with just the + sign) dx = 2 [c - bt + t2 ](t2-c)-2 dt OK, now the next step is to get R as a function of t! This does not look good because we want to talk about . Write it out N = a t4 - 2bt3 + (b2 + 2ac)t2 - 2bc t + ac2 How on earth is this going to "simplify" our integral? We will have square root of the above in the denominator. Maple says this does not factor. Write out 6 terms like so N = a t4 - 2bt3 + b2t2 + 2act2 - 2bc t + ac2 Replace a by α2 and = α : N = α2 t4 - 2bαt3 + b2t2 + 2α2ct2 - 2bαc t + α2c2 This was the trick, Here is what Maple says, Then we can say R = (αt2-bt+αc)2/ (c-t2)2 = ± (αt2-bt+αc)/ (c-t2) From above we have dx = 2 [c - bt + t2 ](t2-c)-2 dt = 2 [αc - bt + αt2 ](t2-c)-2 dt = 2(αt2 - bt + αc)(t2-c)-2 dt Then we get the payoff dx/ = ± 2(αt2 - bt + αc)(t2-c)-2 dt // (αt2-bt+αc)/ (c-t2) = ± 2(t2-c)-2 dt // 1/ (c-t2) = ± 2(t2-c)-2 dt * (c-t2) = ∓ 2(t2-c)-2 dt * (t2-c) = ∓ 2(t2-c)-1 dt Once again !Syntax Error, I dx/ = ∓ 2 !Syntax Error, I dt/(t2-c) where t(x) = ( ∓ )/x and I do admit that is a significant simplification. Let's now just look up that integral, Schaum p 64 says So lets take the second one with the log and see what we get with a2 = c !Syntax Error, I dx/ = ∓ 2 !Syntax Error, I dt/(t2-c) = ∓ 2 [ (1/2) ln ] t(x) = ( ∓ )/x xt(x) = ∓ So mult up and down by x to get !Syntax Error, I dx/ = ∓ (1/) ln I am trying to end up with something like this, Pause at 1:30. I have at least made some progress using one of the Euler substitutions, but I have not been able to replicate any of the results that GR7 claims. This is the first time I have ever tried this in my life!