Posted by Subhotosh Khan on September 10, 2002 at 08:42:04:
In Reply to: How does the "adding zero" trick work??? posted by colin on September 10, 2002 at 01:13:29:
: : 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?
: : And THAT's why there are two integrals.
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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]}