"Lo dee-high minus high dee-low over the square of what's below" is my mnemonic. Incidentally, while you never technically need the quotient rule, I find it helpful because it eliminates the need to get a common denominator and add fractions later. More calculus now = less algebra later.
I always think deriviative of the thing in the bracket (which in this case is found by the quotient rule ) multiplied by what you would normally do in differentiation. Ok not worded very well. I end up with -1/((1+x^2)*sqrt((1-x)/(1+x))). Hmm maybe too many brackets spoil the broth
the thing "inside the root" is: (1-x)/(1+x), which has the derivative [(1+x)(-1) - (1-x)(1)]/(1+x)^2 = -2/(1+x)^2.
as HallsofIvy pointed out, (√u)' = 1/(2√u) = (1/2)√(1/u), which in this case becomes (1/2)√((1+x)/(1-x)).
combining the two the 1/2 and 2 cancel, and we may write (1+x)^2 as √(1+x)^4, making the answer:
-1/(√[(1-x)(1+x)^3]) ( will attempt to put this in tex:
hopefully it will display right), which integrates correctly according to mathematica.
Thanks for the help, I'm kicking myself over the quotient formula as that's what the previous chapter of the book was on, I must have misread it when flicking backwards.
Anyway, I think I have an answer now...
Let
Therefore
And
Therefore
That's giving me the same answer as the book but I'm not sure if I've worked out correctly as while looks right it seems very counter intuitive to me.
Bah, Deveno just typed the whole thing out while I was struggling with LaTeX and answered the question. Thanks very much.