Re: How does the

Posted by Soroban on September 10, 2002 at 14:09:34:

In Reply to: Re: How does the posted by Subhotosh Khan on September 10, 2002 at 08:42:04:

: : : The integral becomes: INT 2u du/(g - cu),
: : : which will require some long division (or some
: : : trick like "adding zero").

: : Can you please explain how to do the trick?

: ***********************************************
: INT [2u/(g - cu)du]

: = 2/c{INT [cu/(g - cu)du]}

: = 2/c{INT [(cu - g + g)/(g - cu)du]} - g + g .... this is addition of '0' you have to use this trick to do partial fractions)/

: = 2/c{INT [g/(g - cu)du - du]}

Nicely explained, Mr. K!
(Did anyone notice? You did a "multiply by one" in the first step.)