Posted by Colin on September 10, 2002 at 17:56:42:
In Reply to: I haven't worked it out ...but posted by Subhotosh Khan on September 10, 2002 at 15:37:34:
: ******************************
: 2/c*int[g/(g-cu)]du
: let g - cu = y
: dy = - c du
: 2/c*int[g/(g-cu)]du
: =-2g/c^2*int[dy/y]... there is your "square" term in the denomenator
: try to be very explicit when working with "unwieldy" integration/differentiation. If skip steps - you are liable to mess up. It takes little more time - but - no body gives points for "quick" wrong answer.
Yes. I see your point! I am constantly re-discovering the truth of what you say.
thanks again.